$\sum_{n=1}^{\infty}\frac{n^p}{n!} \in \: e\mathbb{N}$ Let $p \in \mathbb{N}\setminus\{0\}.$ Show that
$$\sum_{n=1}^{+\infty}\frac{n^p}{n!} \in \: e\mathbb{N}$$
I have already shown the convergence using Stirling's formula.
 A: This can be done in a totally elementary way. Write $(n)_k = n(n-1) \dots (n-(k-1)) = \frac{n!}{(n-k)!}$ for the falling factorial. First, observe that
$$\sum_{n=0}^{\infty} \frac{(n)_k}{n!} = \sum_{n=k}^{\infty} \frac{1}{(n-k)!} = e.$$
Second, prove by induction (or otherwise, e.g. using finite differences) that there exist integers $\left\{ k \atop i \right\}$ such that
$$n^k = \sum_{i=0}^k \left\{ k \atop i \right\} (n)_i.$$
Then
$$\sum_{n=0}^{\infty} \frac{n^k}{n!} = e \sum_{i=0}^k \left\{ k \atop i \right\}.$$
Incidentally, $\left\{ k \atop i \right\}$ are the Stirling numbers of the second kind, but you don't need this to prove the desired result. This sum can be interpreted as ($e$ times) the $k^{th}$ moment of the Poisson distribution with $\lambda = 1$, and the Stirling number formula shows that it is equal to ($e$ times) the Bell number $B_k$; this result is called Dobinski's formula.

Okay, an even easier argument is available: we can do it directly by strong induction. Write
$$S_k = \frac{1}{e} \sum_{n=0}^{\infty} \frac{n^k}{n!}.$$
We have $S_0 = 1$ and in general
$$\begin{align} S_{k+1} &= \sum_{n=0}^{\infty} \frac{n^{k+1}}{n!} \\
 &= \sum_{n=1}^{\infty} \frac{n^k}{(n-1)!} \\
 &= \sum_{n=0}^{\infty} \frac{(n+1)^k}{n!} \\
 &= \sum_{n=0}^{\infty} \frac{1}{n!} \left( \sum_{i=0}^k {k \choose i} n^i \right) \\
 &= \sum_{i=0}^k {k \choose i} S_i \end{align}.$$
This is a recurrence satisfied by the Bell numbers, but you don't need this.
A: You can use the following argument. Define
$$f(x):=e^{e^x}-1=\sum_{n\ge 1}\frac{1}{n!}e^{nx}\,.$$
Observe that the $p$th derivative of the series is
$$f^p(x)=\sum_{n\ge 1}\frac{n^p}{n!}e^{nx}\,,$$
hence your desired sum is simply $f^p(0)$.  To see that is satisfies your conclusion, we use the explicit expression for $f$ and consider the base case $f’(x)=e^xe^{e^x}$ and suppose $f^k(x)=\sum_{j=1}^ka_je^{jx}e^{e^x}$ for some natural numbers $a_j$; then we have
$$f^{k+1}(x)=\sum_{j=1}^ka_j\frac{d}{dx}e^{jx}e^{e^x}$$
$$=\sum_{j=1}^ka_j\left(e^{(j+1)x}e^{e^x}+j e^{jx}e^{e^x}\right)$$
$$=\sum_{j=1}^{k+1}(a_{j-1}+a_jj)e^{jx}e^{e^x}\,,$$
where $a_0:=0$. We quickly conclude that $f^p(x)$ is a sum over natural number multiples of $e^{kx}e^{e^x}$ for all $1\le k\le p$, and hence $f^p(0)\in e\mathbb{N}$, as desired.
