algebra precalculus - Prove $e^{i pi} = -1$

I recently heard that $e^{i \pi} = -1$.

WolframAlpha confirmed this for me, however, I don't see how this works.

Answer

This identity follows from Euler's Theorem,

$$\begin{align} e^{i \theta} = \cos \theta + i \sin \theta, \end{align}$$ which has many proofs. The one that I like the most is the following (sketched). Define $f(\theta) = e^{-i \theta}(\cos \theta + i \sin \theta)$. Use the quotient rule to show that $f^{\prime}(\theta)= 0$, so $f(\theta)$ is constant in $\theta$. Evaluate $f(0)$ to prove that $f(\theta) = f(0)$ everywhere.

Take $\theta = \pi$ for your claim.