Deriving $e^{ix} = 1 + ix + \frac{(ix)^2}{2!}+...$ I know how to prove equation:
$$e = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right)^x$$
How can I now derive the series:
$$e^{ix} = 1 + ix + \frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+...$$
Those two seem very similar to me...
 A: Use the binomial theorem on the following limit:
$$e^z = \lim_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x$$
i.e., 
$$\begin{align} \left( 1 + \frac{z}{x} \right)^x &= 1 + \frac{x}{1!} \frac{z}{x}  + \frac{x (x-1)}{2!} \left (\frac{z}{x} \right )^2 + \frac{x (x-1) (x-2)}{3!} \left (\frac{z}{x} \right )^3 + \ldots \\ &= 1 + \frac{z}{1!} + \frac{z^2}{2!} \left ( 1-\frac{1}{x} \right ) + \frac{z^3}{3!} \left ( 1-\frac{1}{x} \right ) \left ( 1-\frac{2}{x} \right ) + \ldots \\ \end{align} $$
As $x \rightarrow \infty$, the product terms approach $1$.  Therefore
$$\lim_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x = \sum_{k=0}^{\infty} \frac{z^k}{k!} = e^z$$
Now, for $z=i x$,
$$e^{i x} = \sum_{k=0}^{\infty} \frac{(i x)^k}{k!} = 1 + i x + \frac{(i x)^2}{2!} + \ldots$$
A: If we are allowed to use calculus,
let $$z=1+\frac y{1!}+\frac{y^2}{2!}+\frac{y^3}{3!}+\frac{y^4}{4!}+\cdots$$
So, $$\frac{dz}{dy}=0+1+\frac y{1!}+\frac{y^2}{2!}+\frac{y^3}{3!}+\frac{y^4}{4!}+\cdots=z$$
$$\implies \frac{dz}z=dy$$
Integrating both sides, $\log z=y+c$
For $y=0,z=1\implies c=\log1-0=0\implies \log z=y\iff z=e^y$
