I stumbled upon a similar problem and really liked the answers there, so I wondered if there were a general solution for
$$\sum_{k=1}^{\infty}\frac{k^n}{k!}=?$$
Sadly, when I try to apply some of the methods from the answers on the other problem, I fall short of a complete answer.
My attempt:
$$e^x=\sum_{k=1}^{\infty}\frac{x^k}{k!}$$
Differentiate once, then multiply both sides by $x$.
$$xe^x=\sum_{k=1}^{\infty}\frac{kx^k}{k!}$$
Differentiate both sides again, then multiply by $x$.
$$x(x+1)e^x=\sum_{k=1}^{\infty}\frac{k^2x^k}{k!}$$
Repeat this process $n$ times to find the general solution for $\sum_{k=1}^{\infty}\frac{k^n}{k!}$ (substituting $x=1$), however, I cannot see a way to proceed like this.
UPDATE
I've found that Dobinski's formula is directly related, and that $\sum_{k=1}^{\infty}\frac{k^n}{k!}=eB_n$
Where $B_n$ is Bell's numbers.
I'll leave this question open for anyone who wants to attempt the problem using derivatives.
If you compute the series with the very first terms you'll get for the series:
$$S = \sum_{k = 1}^{+\infty}\frac{k^n}{k!}$$
$n = 1 ~~~~~ \to ~~~~~ S = e$
$n = 2 ~~~~~ \to ~~~~~ S = 2e$
$n = 3 ~~~~~ \to ~~~~~ S = 5e$
$n = 4 ~~~~~ \to ~~~~~ S = 15e$
$n = 5 ~~~~~ \to ~~~~~ S = 52e$
$n = 6 ~~~~~ \to ~~~~~ S = 203e$
$n = 7 ~~~~~ \to ~~~~~ S = 877e$
$n = 8 ~~~~~ \to ~~~~~ S = 4140e$
Looking at the coefficient in front of "e" in every term, we recognize the famous succession
$$1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, \cdots $$
Which are the famous Bell's Numbers. Then your series can be compute in this way:
$$S = \sum_{k = 1}^{+\infty}\frac{k^n}{k!} = \mathcal{B}_n\ e$$
Where $\mathcal{B}_n$ is the $n$-th Bell's number. This clearly depends on what $n$ you take initially in the series, then it's easily evaluable.
More here
https://en.wikipedia.org/wiki/Bell_number
http://mathworld.wolfram.com/BellNumber.html