Generating function for $\sum_{n=0}^{\infty} n^{p} x^n$ I am trying to derive the generating function for $H(x,p) = \sum_{n=0}^{\infty} n^p x^n$
I am trying to solve it with the following logic:
(Edited now, trying a new framing)
Base case:
$$H(x,0) = \sum_{n=0}^{\infty} x^n = \dfrac{1}{1-x}$$
And
$$H(x,p) = x \dfrac{\partial}{\partial x} H(x,p-1)$$
Implying
$$H(x,p) = \left(x \dfrac{\partial}{\partial x}\right)^p \dfrac{1}{1-x}$$
 A: Hint. An approach similar to what you are looking for. One may recall the general formula

$$
\sum_{n=0}^\infty f(n)x^n=\frac1{1-x}\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}\:\omega_n\left(\frac{x}{1-x} \right), \quad |x|<1, \tag1
$$ 

with the polynomials $$
\omega_n\left(x \right)=\sum_{k=0}^n \begin{Bmatrix}n\\k\end{Bmatrix}k!x^k \tag2
$$ where $\displaystyle \begin{Bmatrix}n\\k\end{Bmatrix}$ are the Stirling numbers of the second kind.
The identity $(1)$ holds true for a vast class of regular functions $f$. A proof of $(1)$ can be found in Boyadzhiev's paper here.
Applying $(1)$ to $f(x)=x^p$, gives directly

$$
\sum_{n=0}^\infty n^px^n=\frac1{1-x}\:\omega_p\left(\frac{x}{1-x} \right), \quad |x|<1. \tag3
$$

Observe that we have
$$
\left(x \frac{d}{dx} \right)^p\frac1{1-x}=\frac1{1-x}\:\omega_p\left(\frac{x}{1-x} \right).\tag4
$$
A: Here's one way of describing the answer. First, the Stirling numbers of the second kind $S(m, k)$ have the property that
$$n^m = \sum_k S(m, k) (n)_k$$
where $(n)_k = n(n-1) \dots (n-(k-1))$. Second, the generating function of $(n)_k$ turns out to be relatively straightforward to write down: it is
$$\sum_n (n)_k x^n = k! \frac{x^k}{(1 - x)^{k+1}}.$$
You can derive this by repeatedly differentiating $\frac{1}{1 - x}$. It follows that
$$\sum_n n^m x^n = \sum_k S(m, k) k! \frac{x^k}{(1 - x)^{k+1}}.$$
This is in fact equivalent to a claim about how the differential operator $\left( x \frac{d}{dx} \right)^n$ expands, as mentioned in the comments. 
