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One way to do it is based on the fact that $\displaystyle \sum_{k=0}^\infty \frac1{k!} = e$ and one can do ''change of variable'' in series.

Notice that

\begin{align}
\sum_{k=1}^\infty\frac{k^2}{k!} &= \sum_{k=1}^\infty\frac{k}{(k-1)!}&&\text{(cancellation)}\\
&=\sum_{n=0}^\infty\frac{n+1}{n!}&&\text{(change of variable)}\\
&=\sum_{n=0}^\infty\frac{n}{n!}+\sum_{n=0}^\infty\frac{1}{n!}&&\text{(linearity)}\\
&=\left(\sum_{n=1}^\infty\frac{n}{n!}\right)+e
\end{align}

Now the problem has been reduced to an ''easier'' one in the sense that the exponent of the numerator goes down to $1$. We can do a similar simplification again:

\begin{align}
\sum_{n=1}^\infty\frac{n}{n!} & = \sum_{n=1}^\infty\frac{1}{(n-1)!} 
= \sum_{n=0}^\infty\frac{1}{k!} = e
\end{align}

So $\displaystyle \sum_{k=1}^\infty\frac{k^2}{k!} = e+e=2e$.