Prove $e^{i \pi} = -1$ 
Possible Duplicate:
How to prove Euler's formula: $\exp(i t)=\cos(t)+i\sin(t)$ ? 

I recently heard that $e^{i \pi} = -1$.
WolframAlpha confirmed this for me, however, I don't see how this works.

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.
