Euler's formula proof with Calculus I was reading this source here and it provides a proof of Euler's formula using calculus. Although I technically understand the reasoning, I can't quite wrap my head around one particular step: if $f(x)= \cos(x)+i \sin(x)$, then $f'(x)=if(x) \implies f(x)=e^{ix}$. I can kind of get this, as derivating $e^{ix}$ gives $ie^{ix}$, and I know that $e^x$ is the only non-constant function s.t. $f'(x)=f(x)$. Is there a more clear way to think about this step? Can we extend that property of $e^x$ to when it has more in the exponent? And why does it have that property anyways? Would there be a way of obtaining this step without simply knowing that $e^{ix}$ would work?
 A: 
Can we extend that property of $e^x$
   to when it has more in the exponent?   

Yes we can.  


And why does it have that property anyways?  

Good question. This is because complex derivatives work almost exactly the same as you are used to real derivatives working. Take the formula for the complex derivative as an example
$$f'(x)=\lim_{h \to 0} \frac{f(z+h)-f(z)}{h}$$
Where $h$ and $z$ are complex numbers. You can go ahead and plug $e^z$ into this function and find it is complex differentiable. The proof is pretty much exactly the same as in the real case. If you want a more geometric approach then I recommend this video.  


is there a more clear way to think about this step?  

Depends on your definition of clear. I think the proof you provided is far clearer than what I am about to post, but perhaps you are defining clarity as rigor? Here is a proof from Wikipedia that is much more formal and explanatory, but I feel it is less clear. Let me know what you want here and I can adjust.

All complex numbers can be expressed in polar coordinates. Therefore, for ''some'' $r$ and $\theta$ depending on $x$,
  $e^{ix} = r (\cos \theta + i \sin \theta)$
  No assumptions are being made about $r$ and $\theta$; they will be determined in the course of the proof. From any of the definitions of the exponential function it can be shown that the derivative of $e^{ix}$ is $ie^{ix}$. Therefore, differentiating both sides gives
  $i e ^{ix} = (\cos \theta  + i \sin \theta) \frac{dr}{dx} + r (-\sin \theta  + i \cos \theta) \frac{d\theta}{dx}$
  Substituting $r(\cos\theta + i\sin\theta)$ for $e^{ix}$ and equating real and imaginary parts in this formula gives $\frac{dr}{dx} = 0$ and $\frac{d\theta}{dx} = 1$. Thus, $r$ is a constant and $\theta$ is $x+C$ for some constant $C$. The initial values $r(0)=1$ and $\theta(0)=0$ come from $e^{0i}=1$, giving $r=1$ and $\theta = x$. This proves the formula
  $e^{ix} = 1(\cos x +i \sin x) = \cos x + i \sin x$

There is something important to note in this paragraph that you are neglecting; we have to solve for the constant of integration. When we solve $y'=y$ we find that $y=ke^x$ and thus we must have some initial condition to solve for $k$.


Would there be a way of obtaining this step without simply knowing that $e^{ix}$ would work?  

We have to know some properties of $e^x$ to ever prove Euler's Formula. Here we just use the fact that $ke^x$ is the only non-constant solution to $y'=y$, which is a pretty natural condition if you ask me.
A: As a complement to the explanation of @Brevan Ellefsen.
Have a look at the figure below. It explains geometrically that $e^{ix}$, imagined as a turning vector at unit speed, has a unitary "speed vector" turning at the same speed, orthogonal to it.
This illustrates the fact that differentiation is equivalent to a $\pi/2$ shift, i.e., in geometrical terms, a $\pi/2$ rotation.
If you analyse it in a separate way:
$$\begin{cases}\cos'(x)=-\sin(x)=\cos(x+\pi/2)\\ \sin'(x)=-\cos(x)=\sin(x+\pi/2)\end{cases}$$
But this boils down to say that the derivative of $e^{ix}$ is $ie^{ix}$, which is interpretated as $e^{ix}$ "turned by" $\pi/2$.

A: For your last question, consider a differential equation of the form
$f: \mathbb{R} \to \mathbb{C}$ and $\forall x \in \mathbb{R}$ $f'(x) = kf(x)$ where $k \in \mathbb{C}$ and $k \neq 0$. 
(Ah! I will still make use of the fact that $e^{kx}$ is a solution but rather, in a different manner)
Consider the following function $g : \mathbb{R} \to \mathbb{C}$, $g(x) = e^{-kx}f(x)$
Since, both exponential function and $f$ are differentiable, so $g$ is differentiable. Hence $g'(x) = -ke^{-kx}f(x)+e^{-kx}f'(x) = e^{-kx}(f'(x)-kf(x)) = 0$ $\forall x \in \mathbb{R}$. Therefore, $g$ is a constant function. Hence,   $g(x) = e^{-kx}f(x) = c \implies f(x) = ce^{kx}$
A: Complex differentiation follows exactly the same process like with the real argument except that we now regard $i$ as an imaginary constant. 
I am posting a bit different not suitable for teaching but as it appeared to me long time ago, so am ready for a bit of down-voting.
If we wish to work on basis of circular trigonometric functions,
$$f(x)= \cos(x)+i \sin(x)$$
$$f'(x)= -\sin(x)+i \cos(x) = if(x) $$
the above is rotation by $ \pi/2$ in complex plane. And,
$$ f''(x)=-f(x)  \rightarrow f''(x)+f(x) = 0 $$
The above is real rotation of period $2 \pi $ in complex plane.
But if we wish to work on basis of hyperbolic functions we can consider differentiation with respect to $ i x$ instead of real $x$ and apply Chain Rule ( divide by the constant $i$).
$$f(x)= \cosh(x)+i \sinh(x)$$
$$ f'(x)= \cosh(x)-i \sinh(x) $$
which is complex conjugate of $f(x).$ And
$$ f''(x)= -\cosh(x)-i \sinh(x)  =-f(x), \rightarrow  f''(x)+f(x) = 0 $$
which is still the same rotation in the complex plane, argument change notwithstanding.
