Why is $e^{i\pi}= -1$

Question

Why does $ e^{i\pi} = -1 $ ?

I know that this form can be used to for instance act on a bloch's sphere (quantum mechanics) using it as $ e^{i\pi/4} $ will do a $ \frac{\pi}{4} $ rotation on the $x-y$ plane

But I found no answer on the internet.

[EDIT]: Moreover the greats answers, I recommend this really well made video for thoses who are not familiary of the Euler's identity.

[EDIT TO CLARIFY]:
Learning the basics of quantum information, I had to use this matrix: $ \begin{pmatrix} 1 && 0 \\\ 0 & e^{i\pi/4} \end{pmatrix} $ which is a gate named $ T(\psi) $ where $ \psi $ is a simple qubit as $ \begin{pmatrix} \alpha \\\ \beta \end{pmatrix} $.

$ \alpha $ and $ \beta $ are just chances to get the classical qubit's state there, for instance $ \psi = \begin{pmatrix} \alpha \\\ \beta \end{pmatrix} $ gets $ \alpha^2 $ chances to be (as classical state) $ \begin{pmatrix} 1 \\\ 0 \end{pmatrix} $ (denoted $ |0\rangle $) and $ \beta^2 $ to be $ \begin{pmatrix} 0 \\\ 1 \end{pmatrix} $ (denoted $ |1\rangle $) (both $ \alpha $ and $ \beta $ can be non null, this is known as superposition).

To visualize a qubit state we can use the Bloch Sphere
The Bloch sphere

Answer 1

$$e^{ix}=1+(ix)+\dfrac{(ix)^2}{2!}+\dfrac{(ix)^3}{3!}+\dfrac{(ix)^4}{4!}...=1+ix+\dfrac{i^2x^2}{2!}+\dfrac{i^3x^3}{3!}+\dfrac{i^4x^4}{4!}...$$

$$=1+ix-\dfrac{x^2}{2!}-i\dfrac{x^3}{3!}+\dfrac{x^4}{4!}...=\left(1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}...\right)+i\left(x-\dfrac{x^3}{3!}...\right) = \cos x+i\sin x$$

So:$$e^{ix}=\cos x+i\sin x$$

And you know the rest if $x=\pi$

Answer 2

HINT: $$\exp(ix)=\cos(x)+i\sin(x)$$

Answer 3

If $z \in \mathbb{C}$, $e^z$ is defined to be the power series $$\sum_{n=0}^\infty \frac{z^n}{n!}$$ This, and the power series for $\sin x$ and $\cos x$ lead to the identity $$e^{i \theta} = \cos(\theta) + i \sin(\theta)$$ from which $e^{i \pi} = -1$ follows.

Answer 4

We have Euler's Identity $e^{i\theta}=\cos \theta + i \sin\theta$.

So, for $\theta =\pi$ we get $e^{\pi i}=-1$