Why does $\sum_{n=0}^{\infty} \frac{\pi^n}{n!} = e^\pi$?

Question

Why is it that

$$ \sum_{n=0}^{\infty} \frac{\pi^n}{n!} = e^\pi \quad ?$$

I think it has to do with the gamma function, but I'm not sure how that would work.

Answer 1

Why does? It actually doesn't. That would be true if you had the infinite series:

$$\sum_{n = 0}^{+\infty} \dfrac{\pi^n}{n!} = e^{\pi}$$

Yet your sum is finite, and we have in general:

$$\sum_{n = 0}^M \dfrac{a^n}{n!} = \frac{e^a \Gamma (M+1,a)}{M!}$$

Where $\Gamma(\cdot, \cdot)$ denotes the incomplete Euler Gamma function.

Answer 2

It is just the exponential series: Since $$\sum_{n=0}^\infty{x^n \over n!}=\exp(x),$$ replacing $x$ with $\pi$ delivers the desired equation.