$\sum_{n=1}^{\infty} \frac{n^2}{ n!}$ equals

You may write, for $n =2,3,4,...$: $$ \frac{n^2}{ n!}=\frac{n^2-n+n}{ n!}=\frac{n(n-1)}{ n!}+\frac{n}{ n!}=\frac{1}{ (n-2)!}+\frac{1}{ (n-1)!} $$ and use a change of indices in the new infinite sums.

HINT: First do the obvious cancellation:

$$\sum_{n\ge 1}\frac{n^2}{n!}=\sum_{n\ge 1}\frac{n}{(n-1)!}\;.$$

Now consider performing some familiar operations on the well-known power series expansion of $e^x$.

Added: Multiply

$$e^x=\sum_{n\ge 0}\frac{x^n}{n!}$$

by $x$ and differentiate:

$$\frac{d}{dx}\left(xe^x\right)=\frac{d}{dx}\left(\sum_{n\ge 0}\frac{x^{n+1}}{n!}\right)=\sum_{n\ge 0}\frac{(n+1)x^n}{n!}=\sum_{n\ge 1}\frac{nx^{n-1}}{(n-1)!}\;.$$