Consider the generalised infinite summation $$S_{n,m}=m^{n+1} \sum_{k=1}^\infty \frac{k^n}{(m+1)^k}=m^{n+1}\,\mathrm{Li}_{(-n)} \left(\frac{1}{m+1}\right)$$
where $m$ and $n$ are positive integers, and $\mathrm{Li}_{s} \left(z\right)$ is the polylogarithm function.
For increasing $n$ starting at zero Mathematica outputs a sequence of polynomials related to A130850 and the Eulerian numbers.
$(n=0)$, $1$;
$(n=1)$, $1+m$;
$(n=2)$, $2+3m+m^2$;
$(n=3)$, $6+12m+7m^2+m^3$;
etc...
The question is how do go about relating $S_{n,m}$ to the sequence of polynomials identified?
---Added---
To be more specific as suggested by Stephen Clark:
$$m^{n+1}\, \text{Li}_{-n}\left(\frac{1}{m+1}\right)=\sum\limits_{k=0}^n (n-k)!\, \mathcal{S}_{n+1}^{(n-k+1)}\, m^k$$ where $n$ is a non-negative integer and $\mathcal{S}_a^{(b)}$ gives the Stirling number of the second kind?
I was initially thinking about the integer sequence generated from $S_{n,m}$ when $m=1$ and that seems to be divisible by ${n+1}$ when it is prime or a power of $5$.
