TITLE: Closed-form for $ \sum_{n=0}^{\infty} \frac{x^n}{n!}n^p, $ where $p\in\mathbb{N}$
QUESTION [6 upvotes]: Is there a closed form to the sum $$ \sum_{n=0}^{\infty} \frac{x^n}{n!}n^p, $$
where $p$ is a positive integer? I know that for $p = 1$ the series sums to $xe^x$, but not sure if there's a simple expression for $p > 1$. If there isn't a closed form, is this series related to any special functions? Thanks!

REPLY [3 votes]: Note that in general $$\left(x\,\frac{\rm d}{{\rm d}x}\right)^{n} f(x) = \sum_{k=1}^n {n \brace k} \, x^k f^{(k)}(x)$$
which can be derived by noting the general form $$\left(x\,\frac{\rm d}{{\rm d}x}\right)^{n} f(x) = \sum_{k=1}^n a_k(n) \, x^k f^{(k)}(x)$$
and then proceed inductively i.e.
$$\left(x\,\frac{\rm d}{{\rm d}x}\right)^{n+1} f(x) = xf^{(1)}(x) + \sum_{k=2}^n \left\{k \, a_k(n) + a_{k-1}(n)\right\} x^k f^{(k)}(x) + x^{n+1} f^{(n+1)}(x) \\
= \sum_{k=1}^{n+1} \left\{ k \, a_k(n) + a_{k-1}(n) \right\} x^k f^{(k)}(x) \stackrel{!}{=} \sum_{k=1}^{n+1} a_k(n+1) \, x^k f^{(k)}(x) $$
where $a_1(n)=1$, $a_n(n)=1$ and $a_0(n)=a_{n+1}(n)=0$ with $n\in \mathbb{N}$.
Finally check that the Stirling Numbers of Second kind $${n\brace k}=\frac{1}{k!} \sum_{i=0}^k (-1)^{k-i} \binom{k}{i} i^n$$ solve the recurrence $$a_k(n+1)=k \, a_k(n) + a_{k-1}(n) \, .$$
Your special case is $f(x)=e^x$.