I am trying to find a closed form expression for $$S_k = \sum_{n=0}^{\infty} \frac{n^k}{n!}$$
For $k=1$, the answer is clearly $e$. For $k=2$, $$\sum_{n=0}^{\infty} \frac{n^2}{n!} = \sum_{n=1}^{\infty} \frac{n^2}{n!} = \sum_{n=1}^{\infty} \frac{n}{(n-1)!} = \sum_{n=0}^{\infty} \frac{n+1}{n!} = S_1 + e = 2e$$
Following the same reduction steps for $n=3$, I get $5e$ ($S_2 + 2S_1 + e$)
Using my computer to find the next few values, the terms seem to be $e, 2e, 5e, 15e, 52e,203e, 877e, ...$
I don't see any discernible pattern in this, otherwise I could solve this by induction using the reduction steps outlined for $S_2$. Alternatively, if we denote $S_0$ to be $e$:
$$\sum_{n=0}^{\infty} \frac{n^k}{n!} = \sum_{n=1}^{\infty} \frac{n^k}{n!} = \sum_{n=1}^{\infty} \frac{n^{k-1}}{(n-1)!} = \sum_{n=0}^{\infty} \frac{(n+1)^{k-1}}{n!} = \sum_{n=0}^{\infty} \frac{\sum_{r=0}^{k-1} \binom{k-1}{r} n^r}{n!} = \sum_{r=0}^{k-1} \binom{k-1}{r} S_r$$
So $S_k$ follows the recurrence relation: $$S_k = \sum_{r=0}^{k-1} \binom{k-1}{r} S_r$$
I don't know how to solve this recurrence either, because it has a variable number of terms for each next term unlike a linear recurrence.
How can I find the closed form for $S_k$? Does it even exist? I don't really know much about these kinds of sequences (I'm in high school) so if you could also provide some sources where I can look for more information on these kinds of series, that would me much appreciated.