For the proof we need to be familiar with the \(\sin\), \(\cos\) and \(e\) taylor series. Let's review the \(\sin\) taylor series:
$$\sin x = \sum^\infty_{n=0} \frac{(-1)^nx^{2n+1}}{(2n+1)!} = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\ldots$$
Here is the \(\cos\) taylor series:
\[\cos x = \sum^\infty_{n=0} \frac{(-1)^nx^{2n}}{(2n)!} = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}\ldots\]
And finally the \(e\) taylor series:
\[e^x = \sum^\infty_{n=0} \frac{x^{n}}{n!} = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\ldots\]
We can break the \(e\) taylor series into even and odd components:
\[\begin{align} e^x &= \sum_{even \ n's} \frac{x^{n}}{n!} + \sum_{odd \ n's} \frac{x^{n}}{n!} \\ &=\sum^\infty_{n=0} \frac{x^{2n}}{(2n)!} + \sum^\infty_{n=0} \frac{x^{2n+1}}{(2n+1)!}
\end{align}\]
Note that:
\[\begin{align} i^{2n} &= (-1)^n, \text{for n = 0, 1, 2, ...}\\ i^{2n+1} &= i(-1)^n, \text{for n = 0, 1, 2, ...}
\end{align}\]
Let \(x = i\Theta\):
\[\begin{align} e^{i\Theta} &=\sum^\infty_{n=0} \frac{(i\Theta)^{2n}}{(2n)!} + \sum^\infty_{n=0} \frac{(i\Theta)^{2n+1}}{(2n+1)!} \\ &=\sum^\infty_{n=0} \frac{(-1)^{n}(\Theta)^{2n}}{(2n)!} + \sum^\infty_{n=0} \frac{i(-1)^n(\Theta)^{2n+1}}{(2n+1)!}
\end{align}\]
This leads to our final step:
\[\begin{align} &=\sum^\infty_{n=0} \frac{(-1)^{n}(\Theta)^{2n}}{(2n)!} + i\sum^\infty_{n=0} \frac{(-1)^n(\Theta)^{2n+1}}{(2n+1)!}\\ &=\cos(\Theta)+i\sin(\Theta)
\end{align}\]