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.