What's the value of $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$?

The sum is equal to $2e$. First of all, the term $k^2/k!$ may be partly canceled as $k/(k-1)!$. Second, this can be written as $(k-1+1)/(k-1)!$. The term $k-1+1$ is divided to two terms. In the first term, the $k-1$ may be canceled again, giving us $e$. The second term leads to $e$ immediately. So the total sum is $2\exp(1)$.

In a similar way, one may easily calculate the sum even if $k^2$ is replaced by $k^n$, any positive integer power of $k$. The result is always a multiple of $e$.

The value of $T_n := \displaystyle\sum_{k=1}^{\infty} \frac{k^n}{k!}$ is $B_n \cdot e$, where $B_n$ is the $n^{th}$ Bell number.

To see this, note that

$$\begin{align} T_{n+1} = \sum_{k=1}^{\infty} \frac{k^{n+1}}{k!} &= \sum_{k=0}^{\infty} \frac{(k+1)^n}{k!} \\ &= \sum_{k=0}^{\infty} \frac{1}{k!} \sum_{j=0}^n {n \choose j} k^j \\ &= \sum_{j=0}^n {n \choose j} \sum_{k=1}^{\infty} \frac{k^j}{k!} \\ &= \sum_{j=0}^{n} {n \choose j} T_j \end{align}$$

This is precisely the recursion formula that the Bell numbers follow, except that every term here is being multiplied by $e$.

Edit: I was unaware at the time I posted my answer, but the argument I gave goes by the name of Dobiński's formula.

Wolfram Alpha says it is $2e$. Another derivation is to start with $e^x=\sum \limits_{n=0}^\infty \frac{x^n}{n!}$, apply $\frac{d}{dx}x\frac{d}{dx}$ to both sides, and evaluate at $x=1$.