Timothy Chow's answer has a nice application. Let $n,x,y,z$ be natural numbers such that $x^n+y^n-z^n=0$. It follows that $e^{x^n+y^n-z^n}=1=e^i$ and the absurd $$1=(e^{x^n+y^n-z^n})^\pi=e^{i\pi}=-1.$$