TITLE: Infinite Series $\sum\limits_{k=1}^{\infty}\frac{k^n}{k!}$
QUESTION [10 upvotes]: How can I find the value of the sum $\sum_{k=1}^{\infty}\frac{k^n}{k!}$?
for example for $n=6$, we have
$$\sum_{k=1}^{\infty}\frac{k^6}{k!}=203e.$$

REPLY [11 votes]: I thought it might be useful to add the derivation of the formula given in Casteels answer.
The defining equation for Stirling Numbers of the Second Kind is
$$
\newcommand{\stirtwo}[2]{\left\{#1\atop#2\right\}}
k^n=\sum_{j=0}^n\stirtwo{n}{j}\binom{k}{j}\,j!\tag{1}
$$
Therefore,
$$
\begin{align}
\sum_{k=0}^\infty\frac{k^n}{k!}
&=\sum_{k=0}^\infty\sum_{j=0}^n\stirtwo{n}{j}\binom{k}{j}\frac{j!}{k!}\\
&=\sum_{j=0}^n\stirtwo{n}{j}\sum_{k=j}^\infty\frac1{(k-j)!}\\
&=e\sum_{j=0}^n\stirtwo{n}{j}\tag{2}
\end{align}
$$
It is worth noting that the Bell Numbers are
$$
B_n=\sum_{j=0}^n\stirtwo{n}{j}\tag{3}
$$
Thus,
$$
\sum_{k=0}^\infty\frac{k^n}{k!}=e\,B_n\tag{4}
$$