Proof for Euler's Identity using Maclaurin Series

Based on

$\sin x = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{(2 n + 1)!} x^{2 n + 1}$
$\cos x = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{(2 n)!} x^{2 n}$

Refer to Math/Complex Numbers to see the link between trigonometry and exponential functions.

And

$e^{x} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!}$

Suppose $x = i θ$

\begin{aligned} e^{i θ} & = \sum_{n = 0}^{\infty} \frac{(i θ)^{n}}{n!} \\ = 1 + i θ + \frac{(i θ)^{2}}{2!} + \frac{(i θ)^{3}}{3!} + \frac{(i θ)^{4}}{4!} + \dots \\ = \overset{\cos θ}{\overset{⏞}{1 - \frac{θ^{2}}{2!} + \frac{θ^{4}}{4!} - \frac{θ^{6}}{6!} \dots}} + i \overset{\sin θ}{\overset{⏞}{(θ - \frac{θ^{3}}{3!} + \frac{θ^{5}}{5!} - \frac{θ^{7}}{7!} \dots)}} \\ = \cos θ + i \sin θ \end{aligned}

Thus

e^{i θ} = \cos θ + i \sin θ

Substituting $θ$ as $π$

e^{i π} = - 1