<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] 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 $y=0,z=1\implies c=\log1-0=0\implies \log z=y\iff z=e^y$
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