Evaluate $\sum_{k=1}^\infty \frac{k^2}{(k-1)!}$. 
Evaluate $\sum_{k=1}^\infty \frac{k^2}{(k-1)!}$

I sense the answer has some connection with $e$, but I don't know how it is. Please help. Thank you.

I have provided a proof of the general case here.  The proof provided shows
$$
\sum_{k=0}^{\infty} \frac{k^n}{k!} = T_n \cdot e
$$
where $n$ is the $n^{th}$ Bell number.  Yours is the case $n = 3$.  Just for completeness, the formula above is called Dobinski's formula.
