General formula for the power sum $\sum_{k=1}^{n}k^mx^k\quad \forall\; m\in \mathbb{N}$ In my last question, it turns out to be solving the formula of $\sum_{k=1}^{n}k\omega^k$. I am curious if there is a geranal formula for the power sum:
$$\sum_{k=1}^{n}k^mx^k\quad \forall\; m\in \mathbb{N}$$
I have calculated first few formula.
\begin{align}
\sum_{k=1}^{n}x^k&=\frac{x}{x-1}(x^n-1)\\
\sum_{k=1}^{n}kx^k&=\frac{x}{x-1}\left(x^n\left(n-\frac{1}{x-1}\right)+\frac{1}{x-1}\right)\\
\sum_{k=1}^{n}k^2x^k&=\frac{x}{x-1}\left(x^n\left(n^2-\frac{2n-1}{x-1}+\frac{2}{(x-1)^2}\right)-\left(\frac{1}{x-1}+\frac{2}{(x-1)^2}\right)\right)\\
\sum_{k=1}^{n}k^3x^k&=\frac{x}{x-1}\left(x^n\left(n^3-\frac{3n^2-3n+1}{x-1}+\frac{6n-6}{(x-1)^2}-\frac{6}{(x-1)^3}\right)+\left(\frac{1}{x-1}+\frac{6}{(x-1)^2}+\frac{6}{(x-1)^3}\right)\right)
\end{align}
The formula has a common form.
$$
\sum_{k=1}^{n}k^mx^k=\frac{x}{x-1}\left(x^nf_m\left(\frac{1}{x-1}\right)-\left.f_m\left(\frac{1}{x-1}\right)\right|_{n=0}\right)\\
f_m \text{ is a polynomial of } \frac{1}{x-1}
$$
It seems there are some rules for $f_m\left(\frac{1}{x-1}\right)$when $m$ increased. For example:
The constant term coefficient of $f_m\left(\frac{1}{x-1}\right)$ is $\;n^m$.
The $\frac{1}{x-1}$ term coefficient of $f_m\left(\frac{1}{x-1}\right)$ is $\;(n-1)^m-n^m$.
The $\frac{1}{(x-1)^m}$ term coefficient of $f_m\left(\frac{1}{x-1}\right)$ is $\;(-1)^mm!$.
I also find a recursive formula for the power sum.
$$\sum_{k=1}^{n}k^{m+2}x^k=\sum_{k=1}^{n}k^{m+1}x^k+n(n+1)\sum_{k=1}^{n}k^mx^k-\sum_{k=1}^{n}\left(2k\sum_{j=1}^{k}j^mx^j\right)$$
Then the formula of $(m+2)^{th}$ order can always be calculated from previous two formula of $(m+1)^{th}$ order and $m^{th}$ order.
I have calculated the formula up to $6^{th}$ order, but the calculations become extremely complex and tedious when the order increased.
The term $$\sum_{k=1}^{n}\left(2k\sum_{j=1}^{k}j^mx^j\right)$$ will be expanded to all the formula from order $1$ to order $m+1$, and each formula times a polynomial of $\frac{1}{x-1}$.
Is there a general formula like Faulhaber's formula to calculate the coefficients of the formula for this power sum in a easier way?
 A: Recalling the umbral calculus relation:
$$ (x \frac{d}{dx})^m g(x)= \sum_{f=0}^{\infty} { m \brace f} \: x^f \: \frac{d^f}{dx^f} g(x) $$
where ${n \brace m} $ are the Stirling Numbers of the Second Kind.
One then has:
$$ \sum_{k=1}^{n} k^m x^k = \sum_{f=0}^{\infty} { m \brace f} \: x^f \: \frac{d^f}{dx^f} \frac{1-x^{n+1}}{1-x} $$
Despite the sum going to infinity, the sum is finite for integer values $m$. For example the sum $$ \sum_{k=1}^{10} k^4 x^k = \left(\frac{1-x^{11}}{(1-x)^2}-\frac{11 x^{10}}{1-x}\right) x+7 \left(\frac{2 \left(1-x^{11}\right)}{(1-x)^3}-\frac{22 x^{10}}{(1-x)^2}-\frac{110 x^9}{1-x}\right) x^2+6 \left(\frac{6 \left(1-x^{11}\right)}{(1-x)^4}-\frac{66 x^{10}}{(1-x)^3}-\frac{330 x^9}{(1-x)^2}-\frac{990 x^8}{1-x}\right) x^3+\left(\frac{24 \left(1-x^{11}\right)}{(1-x)^5}-\frac{264 x^{10}}{(1-x)^4}-\frac{1320 x^9}{(1-x)^3}-\frac{3960 x^8}{(1-x)^2}-\frac{7920 x^7}{1-x}\right) x^4 $$
which, of course, is equivalent to $$10000 x^{10}+6561 x^9+4096 x^8+2401 x^7+1296 x^6+625 x^5+256 x^4+81 x^3+16 x^2+x $$
