TITLE: General term formula of $S_n=\sum\limits_{i=1}^n{\frac{a^i}{i!}}$?
QUESTION [6 upvotes]: If $S_n=\sum\limits_{i=1}^n{\frac{a^i}{i!}}$, where a is a positive number, then what's the general term formula of $S_n$?

REPLY [4 votes]: We could use first-year calculus to get an answer.  By Taylor's Theorem with the integral form of the remainder, we have
$$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots +\frac{x^n}{n!}+\int_0^x \frac{e^t(x-t)^n}{n!}dt.$$
Thus 
$$S_n=e^a-1-\int_0^a \frac{e^t(a-t)^n}{n!}dt.$$
The expression for $S_n$ can be transformed in various ways, but will remain stubbornly non-elementary.