Euler's Identity

Theorem

 * $e^{i \pi} + 1 = 0$

Proof
Follows directly from Euler's Formula $e^{i z} = \cos z + i \sin z$, by plugging in $z = \pi$:


 * $e^{i \pi} + 1 = \cos \pi + i \sin \pi + 1 = -1 + i \times 0 + 1 = 0$

Also presented as
This result can also be presented as:
 * $e^{i \pi} = -1$

or:
 * $e^{\pi i} = -1$