Solution: We recall that the Taylor's series expansion for $e^x$ is given by \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}. \] Setting $x = \pi $, we get \[ e^\pi = \sum_{n=0}^{\infty} \frac{\pi ^n}{n!}. \] Thus, the value of the infinite sum is \[ \textcolor{blue}{\boxed{ \sum_{n=0}^{\infty} \frac{1}{n!} = e }} \]