$\sum_{i=0}^b \frac{n^i}{i!}$ =?

Question

I am trying to solve the summation $\sum_{i=0}^b \frac{n^i}{i!}$. I know that it´s the Taylor exapnsion of $e^n$ as b $\rightarrow \infty$, but I´m having some trouble with the remainder. Is there a closed form in this case? If not, it would be enough to upper-bound it (not by the obvious $e^n$ but a little bit tighter).