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.

Let's finish it in one line
$$\sum_{k=1}^\infty \frac{(k-2)(k-1)+3(k-1)+1}{(k-1)!}=5e$$
Chris.
