<s>[INST] 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...
 [/INST] 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 $z=i x$,
$$e^{i x} = \sum_{k=0}^{\infty} \frac{(i x)^k}{k!} = 1 + i x + \frac{(i x)^2}{2!} + \ldots$$
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