Simple way to evaluate $f(n)= \sum_{r=0}^{\infty}r^n x^r$ Does anyone know a simpler formula than the one below for calculating values of this function for any positive integer n?
$$f(n)= \sum_{r=0}^{\infty}r^n x^r$$
Here's the derivation for the best formulae I could come up with:
$$ \frac {1}{1-x} = \sum_{r=0}^{\infty} x^r$$
Differentiating n times:
$$ \frac {n!}{(1-x)^{n+1}} = \sum_{r=0}^{\infty}(r+1)(r+2)...(r+n) x^r=\sum_{r=0}^{\infty} \frac 1r r^{(n+1)} x^r$$
Where $r^{(n)}$denots rising factorial. Expanding rising factorial in terms of Unsigned Stirling numbers of the first kind:
$$ \frac {n!}{(1-x)^{n+1}} = \sum_{r=0}^{\infty}\frac 1r \sum_{k=1}^{n+1}c(n+1,k)r^kx^r=\sum_{r=0}^{\infty} \sum_{k=0}^{n}c(n+1,k+1)r^kx^r \\ \therefore \sum_{r=0}^{\infty}c(n+1,n+1)r^nx^r=\frac {n!}{(1-x)^{n+1}}-\sum_{r=0}^{\infty} \sum_{k=0}^{n-1}c(n+1,k+1)r^kx^r $$
Since $c(n+1,n+1)=1$, $$ f(n)=\frac {n!}{(1-x)^{n+1}}-\sum_{k=0}^{n-1}c(n+1,k+1)f(k)$$ This is a recurrence relation with $f(0)=\frac {1}{1-x}$ and thus should be solvable but I don't have the skill to do so.
Also, this function can be represented as a polylogarithm and this leads to a closed-form but still complicated solution in terms of Stirling numbers of the second kind:
$$f(n)=Li_{-n}(x)=\sum_{r=0}^{\infty} (-1)^{n+r}S(n+1,r+1)\frac{r!}{(1-x)^{r+1}}$$
Any help solving the recurrence or exploiting properties of the polylogarithm would be very much appreciated.
ALSO solving the recurrence would lead to a relation between Stirling numbers of the first and second kinds and a different representation of the polylogarithm.

A slightly different approach is as follows:
Let
$$
S(t) = \sum_{r=0}^\infty e^{rt}x^r = \sum_{r=0}^\infty (e^t x)^r = \frac{1}{1-e^t x},
$$
for sufficiently small $x$. Then
$$
S^{(n)}(t) = \frac{\partial^n}{\partial t^n} \left[ \frac{1}{1-e^t x} \right] = \frac{e^t x}{(1-e^t x)^{n+1}} \sum_{k=0}^{n-1} A(n,k+1) (e^t x)^k,
$$
where $A(n,k+1)$ are the Eulerian numbers. Now, set $t=0$ to obtain
$$
S^{(n)}(0) = f(n) = \sum_{r=0}^\infty r^n x^r = \frac{x}{(1-x)^{n+1}} \sum_{k=0}^{n-1} A(n,k+1) x^k = \frac{1}{(1-x)^{n+1}} \sum_{k=0}^{n-1} A(n,k+1) x^{n-k}.
$$
I don't know if this should be considered 'simpler' than your formula, but at least it rids you of recurrences and infinite sums.
