Is there a close form of $ S_n =\sum_{k=0}^{\infty}\frac{k^n}{k!}$

Question

I want to fine the close form of the following sequence

$$S_n =\sum_{k=0}^{\infty}\frac{k^n}{k!}$$

here is my attempt ,

$$S_0 = \sum_{k=0}^{\infty}\frac{1}{k!} =e~~~\text{and }~~~S_1 = \sum_{k=1}^{\infty}\frac{1}{(k-1)!} = e$$ and $$S_2 = \sum_{k=1}^{\infty}\frac{k}{(k-1)!} =\sum_{k=0}^{\infty}\frac{k+1}{k!} =S_0+S_1 =2e$$
More generally, by Binomial formula I have $$S_{n+1} =\sum_{k=0}^{\infty}\frac{k^{n+1}}{k!}=\sum_{k=0}^{\infty}\frac{(k+1)^{n}}{k!}=\sum_{j=0}^{n}{n\choose j}\sum_{k=0}^{\infty}\frac{k^j}{k!}=\sum_{j=0}^{n}{n\choose j}S_j$$

Icannot continue from here. can some tell me what to do?

Answer 1

$$ S_n =\sum_{k=0}^{\infty}\frac{k^n}{k!} = B_n e $$

where $B_n$ is the $n$-th Bell Number.

Answer 2

What is seems is that $$S_n=e\, B_n$$ where appears Bell numbers.