How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$? Could you provide a proof of Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
 A: Let $\mathbf{A}$ be an $n \times n$ matrix.  Recall that the system of differential equations
$$\mathbf{x}' = \mathbf{Ax}$$
has the unique solution $\mathbf{x} = e^{\mathbf{A}t} \mathbf{x}(0)$, where $\mathbf{x}$ is a vector-valued differentiable function and $e^{\mathbf{A}t}$ denotes the matrix exponential.  In particular, let $\mathbf{J} = \left[ \begin{array}{cc} 0 & 1 \\\ -1 & 0 \end{array} \right]$.  Then the system of differential equations
$$x' = y, y' = -x$$
with initial conditions $x(0) = 1, y(0) = 0$ has the unique solution $\left[ \begin{array}{cc} x \\\ y \end{array} \right] = e^{\mathbf{J}t} \left[ \begin{array}{cc} 1 \\\ 0 \end{array} \right]$.  On the other hand, the above equations tell us that $x'' = -x$ and $y'' = -y$, and we know that the solutions to this differential equation are of the form $a \cos t + b \sin t$ for constants $a, b$.  By matching initial conditions we in fact find that $x = \cos t, y = \sin t$.  Now verify that on vectors multiplying by $\mathbf{J}$ has the same effect as multiplying a complex number by $i$, and you obtain Euler's formula.
This proof has the following attractive physical interpretation: a particle whose $x$- and $y$-coordinates satisfy $x' = y, y' = -x$ has the property that its velocity is always perpendicular to and has proportional magnitude to its displacement.  But from physics lessons you know that this uniquely describes particles which move in a circle. 
Another way to interpret this proof is as a description of the exponential map from the Lie algebra $\mathbb{R}$ to the Lie group $\text{SO}(2)$.  Euler's formula generalizes to quaternions, and this in turn can be thought of as describing the exponential map from the Lie algebra $\mathbb{R}^3$ (with the cross product) to $\text{SU}(2)$ (which can then be sent to $\text{SO}(3)$).  This is one reason it is convenient to use quaternions to describe 3-d rotations in computer graphics; the exponential map makes it easy to interpolate between two rotations.
Edit:  whuber's answer reminded me of the following excellent graphic.

This is what is happening geometrically in whuber's answer, and is essentially what happens if you apply Euler's method to the system of ODEs I described above.  Each step is a plot of the powers of $1 + \frac{i}{N}$ up to the $N^{th}$ power.
A: One could provide answers based on a wide range of definitions of $\exp$, $\cos$, and $\sin$ (e.g., via differential equations, power series, Lie theory, inverting integrals, infinite sums, infinite products, complex line integrals, continued fractions, circular functions, and even Euclidean geometry) as well as offering Euler's formula up as a tautology based on a definition.  But let's consider where in one's education this question arises: it's usually well before most of these other concepts are encountered.  The complex numbers have just been introduced; the Argand diagram is available; $\exp(x)$ is most likely defined as the limiting value of $(1 + x/n)^n$, and $\cos$ and $\sin$ are defined as circular functions (that is, as arclength parameters for coordinates of the unit circle).  So the question deserves a response in this context using mathematics accessible to someone at this level.
Accordingly, I propose that we interpret $\exp(i x)$ as the limit of $(1 + i x/n)^n$, because at least one understands how to compute the latter (as iterated multiplication of 1 by $1 + i x/n$), one at least vaguely intuits what the limit of a sequence of points in the complex plane might mean, and one has learned that $\cos(\theta) + i \sin(\theta)$ is quite literally the complex number plotted at $\left( \cos(\theta), \sin(\theta) \right)$ in the plane.
It seems natural to evaluate the limit by looking at its modulus and argument separately.  The modulus is relatively easy: because $|1 + i x/n| = \sqrt {  1 + (x/n)^2  }$, the modulus of its $n$th power equals $\left( 1 + (x/n)^2 \right) ^{n/2}$.  We can exploit the one limit assumed to be known:
$\left( 1 + (x/n)^2 \right) ^{n/2} = \left( \left( 1 + x^2 / n^2 \right) ^ {n^2} \right) ^ {1/{2n}} \simeq \left( \exp ( x^2 ) \right) ^ {1/{2n}} \to 1$.
(There's some hand-waving in the last two steps.  This is characteristic of limiting arguments at this level.  Those of you who can see where the rigor is lacking also know exactly how to supply the missing steps.)  Whence, whatever $\exp( i x )$ might be, we deduce it should lie on the unit circle.
Now we turn our attention to the argument of the limit.  The sequence of arguments
$\left( 1 + (x/n)^2 \right) ^{1/2}, \left( 1 + (x/n)^2 \right) ^{2/2}, \ldots, \left( 1 + (x/n)^2 \right) ^{k/2}, \ldots$
obviously is non-decreasing, because it's a geometric sequence with multiplier of 1 or greater.  That is, each successive multiplication by $1 + i x / n$ is expanding an original right triangle with vertices at (0,0), (1,0), and (1, $x/n$), but in the limit the amount of expansion is reduced to unity, as we have seen.  In the Argand diagram we're just laying out similar copies of this original right triangle, placing one leg of each new copy along the hypotenuse of the previous one.  In the limit, the length of the small leg (originally $x/n$) therefore remains constant.  These observations imply that near the limit, we can take all these $n$ little triangles to be essentially congruent, whence the length of the path traced out by the succession of images of the small leg must be extremely close to $n (x/n) = x$.  This pins down where on the circle $\exp( i x )$ must lie: it is the point reached by traveling (signed) distance $x$ counterclockwise along the circle beginning at (1,0).  To anyone exposed to the definition of $\sin$ and $\cos$ in terms of circular functions, it is immediate that the coordinates of this limiting location are $\left( \cos(x), \sin(x) \right)$ and we are done.
A: How about the Laplace transform of
$$\exp(i t)=\cos(t)+i\sin(t)$$   
Let's evaluate the Laplace transform of the two sides of the formula:  
$$\frac{1}{s-i}=\frac{s}{s^2+1}+\frac{i}{s^2+1}=\frac{s+i}{s^2+1}$$  
Now, let's multiply both sides by $s-i$ :  
$$1=\frac{(s-i)(s+i)}{s^2+1}=\frac{s^2-i^2}{s^2+1}=\frac{s^2+1}{s^2+1}=1$$  
Voilà
A: I've noticed that several of the above answers are heuristic in that they use/operate on the complex exponential in an attempt to define it. This would be like looking up a word in a dictionary and finding the word itself in the definition. Until the complex exponential is defined the expression $ e^{iy}$ has no meaning. One cannot "prove" euler's identity because the identity itself is the DEFINITION of the complex exponential.
So really proving euler's identity amounts to showing that it is the only reasonable way to extend the exponential function to the complex numbers while still maintaining its properties. What follows is basically a more detailed version of what has been summarized above by asmaire. We want to define $$f\left(z\right)=e^z$$ in such a way that it satisfies the following properties: $$ f'\left(z\right) = f\left(z\right) , \quad f\left(x+ 0i\right) = e^x$$  In other words we want it to be its own derivative and we would like it to reduce to the regular exponential function when the exponent is purely real. Lets explore what properties such a function would have to have. 
To make things easier write $f\left(z\right) = f\left(x+iy\right) = u + iv$ with $ u = u\left(x,y\right) , v = v\left(x,y\right)$.
By the Cauchy-Riemann equations we know that $$f'\left(x+iy\right) = u_x + i v_x = v_y + i \left(-u_y\right) = u + iv.$$ where the rightmost equality comes from the fact that$ f' = f$. Equating real/imaginary parts we see that $$u\left(x,y\right) = u_x\left(x,y\right)$$ $$ v\left(x,y\right) = v_x\left(x,y\right)$$ for all x,y. The general solutions to these equations are $$u\left(x,y\right) = a\left(y\right)e^x$$ $$ v\left(x,y\right)= b\left(y\right)e^x $$ Looking at the other constraint we have $$ e^x + 0i = e^ x = f\left(x+0i\right) = u\left(x,0\right) + i v\left(x,0\right) = a\left(0\right)e^x + i b\left(0\right)e^x$$ This gives our "initial conditions" $$ a\left(0\right) = 1 ,\  b\left(0\right) = 0.$$ 
Going back to the C-R equations we have $$ u_x = v_y  \implies a\left(y\right)e^x = b'\left(y\right)e^x$$ $$v_x = - u_y \implies a'\left(y\right)e^x = -b\left(y\right)e^x$$ Giving the system $$a = b'$$ $$- b = a'$$ Which we can cleverly write as $$  \vec{x}' = \begin{bmatrix} a'\left(y\right) \\  b' \left(y\right) \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}  \begin{bmatrix} a\left(y\right) \\ b\left(y\right) \end{bmatrix} = A \vec{x} $$  It turns out the solution to a linear system like this is given by the matrix exponential $$ \vec{x}\left(y\right) = e^{Ay} * \vec{x_0} $$ 
where $  \vec{x_0} =  \begin{bmatrix} a\left(0\right) \\ b\left(0\right) \end{bmatrix} =  \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $ e^{Ay} = \displaystyle∑_{k=0}^{\infty} \frac{A^ky^k}{k!} $ and$A^k$ denotes matrix exponentiation. 
Note that $ A^2 = -I$ so that $ A^3 = - A, A^4 = I, A^5 = A $ etc. 
This gives $$ e^{Ay} = \displaystyle∑_{k=0}^{\infty} \frac{A^ky^k}{k!} = \displaystyle∑_{even k} \frac{A^ky^k}{k!} +  \displaystyle∑_{odd k} \frac{A^ky^k}{k!}$$
$$ = \displaystyle∑_{k=0}^{\infty} \frac{\left(-1\right)^k Iy^{\left(2k\right)}}{2k!} +   \displaystyle∑_{k=0}^{\infty} \frac{\left(-1\right)^kAy^{\left(2k+1\right)}}{\left(2k+1\right)!}$$ $= \begin{bmatrix} ∑_{k=0}^{\infty} \frac{\left(-1\right)^k y^{\left(2k\right)}}{2k!} & 0 \\ 0 & ∑_{k=0}^{\infty} \frac{\left(-1\right)^k y^{\left(2k\right)}}{2k!}\end{bmatrix} +  \begin{bmatrix} 0 & - ∑_{k=0}^{\infty} \frac{\left(-1\right)^ky^{\left(2k+1\right)}}{\left(2k+1\right)!} \\ ∑_{k=0}^{\infty} \frac{\left(-1\right)^ky^{\left(2k+1\right)}}{\left(2k+1\right)!} & 0 \end{bmatrix} $ 
$=   \begin{bmatrix}  ∑_{k=0}^{\infty} \frac{\left(-1\right)^k y^{\left(2k\right)}}{2k!}  & - ∑_{k=0}^{\infty} \frac{\left(-1\right)^ky^{\left(2k+1\right)}}{\left(2k+1\right)!} \\ ∑_{k=0}^{\infty} \frac{\left(-1\right)^ky^{\left(2k+1\right)}}{\left(2k+1\right)!} &  ∑_{k=0}^{\infty} \frac{\left(-1\right)^k y^{\left(2k\right)}}{2k!} \end{bmatrix}= \begin{bmatrix} \cos \left(y\right) & - \sin \left(y\right) \\ \sin \left(y\right) & \cos \left(y\right) \end{bmatrix}$ 
As mentioned above, multiplying by the initial conditions vector gives us our solution : $$ \begin{bmatrix} a\left(y\right) \\ b\left(y\right) \end{bmatrix} =  \begin{bmatrix} \cos \left(y\right) & - \sin \left(y\right) \\ \sin \left(y\right) & \cos \left(y\right) \end{bmatrix}  \begin{bmatrix} 1 \\ 0 \end{bmatrix} =  \begin{bmatrix} \cos \left(y\right) \\ \sin \left(y\right) \end{bmatrix}$$ We finally arrive at $$f\left(x+iy\right) = u\left(x,y\right) + i v\left(x,y\right) = e^x\cos \left(y\right) + i e^x\sin \left(y\right).$$ In other words, if we want the complex exponential to naturally generalize the real exponential then
we must DEFINE it as $ e^{x+iy} = e^x\left(\cos \left(y\right) + i \sin \left(y\right)\right)$.
A: Proof:
Consider the function $f(t) = e^{-it}(\cos t + i \sin t)$ for $t \in \mathbb{R}$. By the product rule 
\begin{eqnarray}
f^{\prime}(t) = e^{-i t}(i \cos t - \sin t) - i e^{-i t}(\cos t + i \sin t) = 0
\end{eqnarray}
identically for all $t \in \mathbb{R}$. Hence, $f$ is constant everywhere. Since $f(0) = 1$, it follows that $f(t) = 1$ identically. Therefore, $e^{it} = \cos t + i \sin t$ for all $t \in \mathbb{R}$, as claimed.
A: An easy way out to answer the question is when we try to extend the definition of exponential to complex plane in a "nice" way (read "nice" as holomorphic) we then end up with this definition. And as you are probably aware there is only at most one such extension. So if we want to define our exponential in the complex plane such that the exponential function is holomorphic, and it matches with the conventional exponential function on the real line, we end up with $e^{it} = \cos(t) + i \sin(t)$. 
All the other answers I think are circular in some sense. 
For instance, how do we know that we can expand $e^{i x}$ as a power series. All we know is that we can expand $e^{x}$ as a power series when $x$ is real. We do not know apriori that the expansion can be carried forward if $x$ happens to be complex.
A: "Combining" the answers from Qiaochu Yuan and Isaac one can also directly evaluate 
$\begin{bmatrix}x \\ y \end{bmatrix} = e^{\mathbf{J}t} \begin{bmatrix}1 \\ 0 \end{bmatrix}$
with $\mathbf{J}=\begin{bmatrix}0&1 \\ -1&0 \end{bmatrix}$ (note that $\mathbf{J}^2 = - \begin{bmatrix}1&0 \\ 0&1 \end{bmatrix}$) as follows:
$$
e^{\mathbf{J}t} \begin{bmatrix}1 \\ 0 \end{bmatrix} = \left( \mathbf{J}^0 + \frac{t}{1!} \mathbf{J}^1 + \frac{t^2}{2!} \mathbf{J}^2 + \frac{t^3}{3!} \mathbf{J}^3 + \frac{t^4}{4!} \mathbf{J}^4 + \cdots \right) \begin{bmatrix}1 \\ 0 \end{bmatrix} \\ = \left( \begin{bmatrix}1&0 \\ 0&1 \end{bmatrix} + t \begin{bmatrix}0&1 \\ -1&0 \end{bmatrix} - \frac{t^2}{2!}\begin{bmatrix}1&0 \\ 0&1 \end{bmatrix}-\frac{t^3}{3!}\begin{bmatrix}0&1 \\ -1&0 \end{bmatrix}+ \frac{t^4}{4!}\begin{bmatrix}1&0 \\ 0&1 \end{bmatrix}+\cdots\right)\begin{bmatrix}1 \\ 0 \end{bmatrix} \\ = \begin{bmatrix} 1 + 0 - \frac{t^2}{2!} - 0 + \frac{t^4}{4!} + \cdots \\ 0 - t - 0 + \frac{t^3}{3!} + 0 + \cdots   \end{bmatrix} \\ = \begin{bmatrix} \cos(t) \\ - \sin(t) \end{bmatrix}
$$
The result is a parametrization of a helix curve. That means Eulers formula simply shows how one can parametrize a helix using the exponential function. 
The result can also be written like
$$
e^{\mathbf{J}t} \begin{bmatrix}1 \\ 0 \end{bmatrix} = \left( \cos(t) + \mathbf{J} \sin(t) \right)\begin{bmatrix} 1 \\ 0 \end{bmatrix}
$$
making the connection to the traditional form of Eulers formula manifest.
It should be noted that with $\mathbf{K}=\begin{bmatrix}0&1 \\ 1&0 \end{bmatrix}$ the following relation can be proved in the same way:
$$
e^{\mathbf{K}t} \begin{bmatrix}1 \\ 0 \end{bmatrix} = \left( \cosh(t) + \mathbf{K} \sinh(t) \right)\begin{bmatrix} 1 \\ 0 \end{bmatrix}
$$
A: Let $f(x) = e^{ix} - \cos(x) - i \sin(x)$. Now $f'' + f = 0$,  $f(0)=0$, $f'(0)=0$ and hence $f(x) = 0$.
A: Let $y=\cos \phi+i\sin \phi$ $...(1)$
Differentiating both sides of equation (1) with respect to $\phi$, we get, 
$\frac{dy}{d\phi}=-\sin \phi+i\cos \phi$
$\implies \frac{dy}{d\phi}=i(\cos \phi-\frac{1}{i}\sin \phi)$
$\implies \frac{dy}{d\phi}=i(\cos\phi+i\sin \phi)$
$\implies \frac{dy}{d\phi}=iy$
$\implies\frac{1}{y}dy=id\phi$ $...(2)$
Integrating both sides of equation (2), we get,
$\int\frac{1}{y}dy=\int id\phi$
$\implies \ln(y)=i\phi+c$ $...(3)$
Substituting $\phi=0$ in equation (1), we get, 
$y=\cos 0+i\sin 0$
$\implies y=1$
Substituting $\phi=0$ and $y=1$ in equation (3) we get,
$\ln(1)=c$
$\implies c=0$
Substituting $c=0$ in eqaution (3) we get, 
$\ln(y)=i\phi$
$e^{i\phi}=y$
$\therefore e^{i\phi}=\cos \phi+i\sin \phi$
A: Hint $ $ Both $\:\rm e^{ix}\:$ and $\:\rm cos(x) + i \; sin(x) \:$ are solutions of $\;\rm y' = i \; y,\;\; y(0) = 1, \;$  so they are equal by the uniqueness theorem. Alternatively, bisect into even & odd parts the power series for $\,\rm e^{ix},\,$  i.e.
$$\begin{align}
\rm f(x) \ \ &=\ \ \rm\frac{f(x)+f(-x)}{2} \;+\; \frac{f(x)-f(-x)}{2} \\[.5em]
\Rightarrow\ \ \ \rm e^{ix} \ \ &=\ \ \rm\cos(x) \ +\ i \:\sin(x)  \end{align}$$
Remarks 1.$\;$ Uniqueness theorems provide powerful tools for proving equalities for functions that satisfy certain nice differential or difference (recurrence) equations. This includes a large majority of functions encountered in theory and practice. Such ideas lie behind algorithms implemented in computer algebra systems, e.g. search the computational literature using the terms "D-finite" and/or "holonomic" functions. 
For a less trivial but still easy example of this technique see my recent post which proves the identity
$$\rm \frac{sinh^{-1}(x)}{\sqrt{x^2+1}} \  = \ \ \sum_{k=0}^\infty\ (-1)^k \frac{(2k)!!}{(2k+1)!!} \: x^{2k+1}$$ 
For a simple discrete example see  here where we remark that $\rm\; 13 \:|\: 3^{n+1} + 4^{2n-1} =: f_n \;$ follows from $\;\rm f_2 = f_1 = 0 \;$ and the (obvious) fact that $\;\rm f_n \;$  satisfies a monic 2nd order linear recurrence. Namely let $\;\rm S\; f_n := f_{n+1} \;$ be the shift operator. Then $\;\rm S - 3 \;$ kills $\rm\; 3^{n+1} \;$ and $\;\rm S - 16 \;$ kills $\;\rm 4^{2n-1} \;$ therefore $\;\rm (S-3)\:(S-16) = S^2 - 19\: S + 48 \;$ kills their sum $\;\rm f_n \:,\;$ i.e. $\;\rm f_{n+2} = 19\; f_{n+1} - 48\; f_n\;$. So $\:\rm mod\:13:\  f_2 = f_1 = 0  \;\Rightarrow\; f_3 = 19\: f_2 - 48\: f_1 = 0 \;\Rightarrow f_4 = 0 \Rightarrow f_5 = 0 \Rightarrow \cdots\Rightarrow\; f_n = 0\: $. So $\:0\:$ is the unique solution of the above recurrence that satisfies the intial conditions $\;\rm f_2 = f_1 = 0. \;$ This is simply an obvious special case of the uniqueness theorem for difference equations (recurrences). Once again, by invoking a uniqueness theorem, we have greatly simplified the deduction of an equality.  See this answer for a simple operator-theoretic generalization of the above method.
Notice that, above, we don't need to know the precise recurrence relation. Rather, we need only know a bound on its degree, so that we know how many initial terms are needed to determine the solution uniquely. In practice, as above, one can often easily derive simple upper bounds on the degree of the recurrence or differential equation - which makes the method even more practical.
2. $\;$ Generalizing the above bisection into even and odd parts, one can employ n'th roots of unity to take arbitrary n-part multisections of power series and generating functions. They often prove handy, e.g.
Exercise $\;$ Give elegant proofs of the following
$\quad\quad\rm\displaystyle sin(x)/e^{x} \quad\:$ has every $\rm 4k\;$'th term zero  
$\quad\quad\rm\displaystyle cos(x)/e^{x} \quad$ has every $\rm 4k+2\;$'th term zero
See the posts in this thread for various solutions and more on multisections.
A: Consider asking the question what is the value of cos(ix) in polar form.
That is find A(x) and B(x) such that
$$\cos(ix)=A+iB$$
Differentiating the equality a couple of times 
$$-i\sin(ix)=A'+iB'$$
$$\cos(ix) = A" +iB"$$
Thus $$A"=A$$ $$B"=B$$
The solution for A or B is $C_1e^x+C_2e^{-x}$. For arbitrary $C_1$ and $C_2$. Something very-clear at the time the equation was developed. Both the question and the solution of the differential $A"=A$ where known at the time the equation was developed.
$$\cos(ix)=C_1e^x+C_2e^{-x} +i[C_3e^x+C_4e^{-x}]$$
$$\cos(i(-ix))=\cos(x)=C_1e^{-ix}+C_2e^{ix} +i[C_3e^{-ix}+C_4e^{ix}]$$
$$\cos(x)=C_1e^{-ix}+C_2e^{ix} +i[C_3e^{-ix}+C_4e^{ix}]$$
$$-sin(x)=C_1(-i)e^{-ix}+C_2(i)e^{ix} +i[C_3(-i)e^{-ix}+C_4(i)e^{ix}]$$
Applying BCs $\cos(0)=1$ and $\sin(0)= 0$ one finds that $C_1=C_2=1/2$ and $C_3=C_4=0$ fit the conditions. 
Hence:
$$\cos(x)=\frac{e^{-ix}+e^{ix}}{2}$$
$$-\sin(x)=\frac{-ie^{-ix}+ie^{ix}}{2}$$ 
$$-i\sin(x)=\frac{e^{-ix}-e^{ix}}{2}$$
Thus
$$\cos(x)+i\sin(x)=e^{ix}$$
More on the BCs. 
$$cos(0)=[C_1+C_2] +i[C_3+C_4]=1$$
$$[C_1+C_2]=1, [C_3+C_4]=0$$
$$-sin(0)=i[C_2-C_1] +[C_3-C_4]=0$$
$$[C_2-C_1]=0, [C_3-C_4]=0$$
Thus $C_1=C_2=1/2$ and $C_3=C_4=0$ fit the conditions. (I cannot demonstrate uniqueness for solution for $C_n$. In so far I accept the uniqueness because the same procedure apply to $sin(ix)$ and $e^{ix}$ yield the same result: Euler's equation.)    
An interesting example:
$$Atan(ix)= A+Bi$$
Taking derivative 
$$ \frac{i}{1-x^2}= A'+B'i$$
Thus
$$ \frac{i}{1-x^2}= B'i$$
$$ \frac{1}{1-x^2}= B'$$ 
solving for B one finds
$$ \frac{1}{2}Ln[\frac{x+1}{1-x}]= B$$
=>
$$ Atan(ix)=\frac{i}{2}Ln[\frac{x+1}{1-x}]= iB$$
Substituting x=-iz, becomes
$$ Atan(z)=\frac{i}{2}Ln[\frac{-iz+1}{1+iz}]$$
which is equivalently expressed as 
$$ Atan(z)=\frac{1}{2i}Ln[\frac{1+iz}{1-iz}]$$ 
A: Well this question actually boils down to "How is the complex exponential defined?"
Here is my view of this problem:
Let
$$f(x+iy)= e^{x}(\cos(y)+i\sin(y) ) \,.$$
Then $f$ has continuous partial derivatives $f_x$ and $f_y$, and verifies the Cauchy-Riemann equations, thus it is analytic.
Moreover, for any $z_1,z_2 \in {\mathbb C}$ we have 
$$f(z_1+z_2)=f(z_1) f(z_2) \,.$$
Last but not least $f(x)=e^x$ for all $x \in {\mathbb R}$.
In particular we showed that $e^x$ can be extended to an analytic complex function, and the theory tells us that such extension is unique.
Now, since $f(z)$ is the unique analytic extension of $e^x$ to the complex plane, and it also satisfies the exponential relation $f(z_1+z_2)=f(z_1) f(z_2)$, we call this function $e^z$. 
A: Assuming you mean $e^{ix}=\cos x+i\sin x$, one way is to use the MacLaurin series for sine and cosine, which are known to converge for all real $x$ in a first-year calculus context, and the MacLaurin series for $e^z$, trusting that it converges for pure-imaginary $z$ since this result requires complex analysis.
The MacLaurin series:
\begin{align}
\sin x&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots
\\\\
\cos x&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots
\\\\
e^z&=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots
\end{align}
Substitute $z=ix$ in the last series:
\begin{align}
e^{ix}&=\sum_{n=0}^{\infty}\frac{(ix)^n}{n!}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\cdots
\\\\
&=1+ix-\frac{x^2}{2!}-i\frac{x^3}{3!}+\frac{x^4}{4!}+i\frac{x^5}{5!}-\cdots
\\\\
&=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots +i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)
\\\\
&=\cos x+i\sin x
\end{align}
A: My favorite proof involves nothing but basic integration (of course it still secretly relies on complex analysis, but looks extremely explicit and straightforward). In fact, this is based on the historical observation made by Bernoulli, which became the first hint of this famous identity. 
Write 
$$\frac{2i}{x^2+1}=\frac{1}{x-i}-\frac{1}{x+i}.$$
Now integrate both sides just as you would integrate real-valued functions of this form:
$$2i\arctan{x}+c=\log\frac{x-i}{x+i}.$$
Replace $x=\tan(y/2)$ and exponentiate (and I'll replace $e^c$ with just $c$ for simplicity):
$$ce^{iy}=\frac{\sin(y/2)-i\cos(y/2)}{\sin(y/2)+i\cos(y/2)}.$$
After the multiplication of both the denominator and the numerator by $\sin(y/2)-i\cos(y/2)$, the right hand side is simplified to
$$\frac{\sin(y/2)-i\cos(y/2)}{\sin(y/2)+i\cos(y/2)}=\frac{(\sin(y/2)-i\cos(y/2))^2}{\sin(y/2)^2+\cos(y/2)^2}=(\sin(y/2)-i\cos(y/2))^2=-\cos y-i\sin y.$$
Substituting $y=0$, we see that $c=-1$, therefore
$$e^{iy}=\cos y+i\sin y.$$
A: Taking same path as whuber's answer, one could also note that if we define $e^x$ as follows,
$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$
Then it follows that
$$e^{ix}=\lim_{n\to\infty}\left(1+\frac{ix}n\right)^n$$
Putting this into polar form, we find that
$$1+\frac{ix}n=\sqrt{1+\frac{x^2}{n^2}}\left(\cos\left(\arctan\left(\frac xn\right)\right)+i\sin\left(\arctan\left(\frac xn\right)\right)\right)$$
By De Moivre's formula,
$$\left(1+\frac{ix}n\right)^n=\left(1+\frac{x^2}{n^2}\right)^{n/2}\left(\cos\left(n\arctan\left(\frac xn\right)\right)+i\sin\left(n\arctan\left(\frac xn\right)\right)\right)$$
One can then see that
$$\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{n/2}=1$$
$$\lim_{n\to\infty}n\arctan\left(\frac xn\right)=x$$
Thus,
$$e^{ix}=\lim_{n\to\infty}\left(1+\frac{ix}n\right)^n=\cos(x)+i\sin(x)$$
A: I'll copy my answer to this question:

Let
$f(x) = \cos(x) + i\cdot \sin(x)$
Then
$$\begin{align*}\frac{df}{dx} &= -\sin(x) + i \cdot \cos(x)\\
      &= i \cdot f(x)
\end{align*}$$
So that
$$
∫(\frac{1}{f(x)}) df = ∫i \cdot dx\\
\ln(f(x)) = ix  + C\\
f(x) = e^{ix + C} = \cos(x) + i \cdot \sin(x)\\
$$
Since $f(0) = 1, C = 0$, so:
$$
e^{ix} = \cos(x) + i \cdot \sin(x)
$$
A: $\newcommand{\E}{\mathrm e}\newcommand{\I}{\mathrm i}$Notice that $\E^{\I t}$ (a real number raised to an imaginary or complex power) is usually defined to be $a^b = \exp(b \log a)$.
Define $\exp$ by the power series:
$$ \exp z := \sum_{j=0}^\infty \frac{z^j}{j!}
$$
which converges on $\mathbb{C}$ by Cauchy-Hadamard formula.
Then $\cos,\sin$ can be defined by
$$ \cos z := \frac{\exp(\I z) + \exp(-\I z)}2,  \qquad \sin z := \frac{\exp(\I z) - \exp(-\I z)}{2\I}
$$
Since $\exp$ is a real power series, $\exp \overline {z} = \overline{\exp z}$. It follows that, if $t \in \mathbb{R}$, we have $\cos t = \Re \exp \I t,\sin t = \Im \exp \I t$.
It remains to define $\log$ and show that $\log \E = 1$. Using Cauchy products, it can be shown that
$$ \exp(w +z) = \exp w \exp z
$$
Since $\exp$ is positive on $[0,+\infty[$ and $\exp 0 = 1$, the above suggests that $\exp$ is positive on $\mathbb R$. Differentiating gives $\exp\restriction_{\mathbb R}' = \exp\restriction_{\mathbb R} > 0$. Using Mean Value Theorem, $\exp$ is strictly increasing on $\mathbb R$ and hence its inverse $\log := \exp\restriction_{\mathbb R}^{-1}$ can be defined.
It is known that $\exp 1 = \E$ (the proof depends on the definition of $\E$) Hence $\log e = 1$, and $\E^{\I t} = \exp(\I t) = \cos t + \I \sin t$.
