Polylogarithms of negative integer order The polylogarithms of order $s$ are defined by
$$\mathrm{Li}_s (z) = \sum_{k \geqslant 1} \frac{z^k}{k^s}, 
\quad |z| < 1.$$
From the above definition, derivatives for the polylogarithms immediately follow. Here we have
$$\frac{d}{dz} \mathrm{Li}_s (z) = \frac{\mathrm{Li_{s - 1}}(z)}{z}.$$
When $s = 0$ the following series which is geometric results
$$\mathrm{Li}_0(z) = \sum_{k \geqslant 1} z^k = \frac{z}{1 - z}.$$
Starting with the expression for $\mathrm{Li}_0(z)$, repeated application of the derivative rule allows explicit expressions for the polylogarithms to all negative integer orders to be found. The first few are:
\begin{align*}
\mathrm{Li}_{-1} (z) &= \frac{z}{(1 - z)^2}\\
\mathrm{Li}_{-2}(z) &= \frac{z(1 + z)}{(1 - z)^3}\\
\mathrm{Li}_{-3}(z) &= \frac{z(1 + 4z +z^2)}{(1 - z)^4}\\
& \vdots
\end{align*}
All are rational functions in $z$. A closed form expression for $\mathrm{Li}_{-n}(z)$ where $n \in \mathbb{N}$ in terms of either Stirling numbers or Eulerian numbers can be found. In terms of Eulerian numbers $\left\langle n \atop k\right\rangle$ the result is
$$\mathrm{Li}_{-n} (z) = \frac{1}{(1 - z)^{n + 1}} \sum^n_{k = 0} \left\langle n \atop k\right\rangle z^{n - k}, \quad n \geqslant 1.$$
What I wish to know is how can such a result be proved. Of course once one knows what one is looking for, the result can be readily guessed, and I suppose induction used to prove the claim is true for all $n$, but is there a way to prove the result directly?    

Suppose we seek to show that
$$\mathrm{Li}_{-n}(w) = \frac{1}{(1-w)^{n+1}}
\sum_{k=0}^{n} \left\langle {n\atop  k}\right\rangle w^{n-k}.$$
We are done if we can prove that
($\left\langle {n\atop  n}\right\rangle$ is zero)
$$[w^m] \frac{1}{(1-w)^{n+1}}
\sum_{k=0}^{n-1} \left\langle {n\atop  k}\right\rangle w^{n-k}
= m^n.$$
Extracting coefficients we get
$$\sum_{k=0}^{n-1} \left\langle {n\atop k}\right\rangle
{m+k-n + n\choose n}
= \sum_{k=0}^{n-1} \left\langle {n\atop k}\right\rangle
{m+k\choose n}.$$
Start from the bivariate generating function of the Eulerian numbers,
which seems like a reasonable starting point and which is
$$\frac{u-1}{u-\exp((u-1)z)}.$$
Now introduce
$${m+k\choose n} = [v^n] (1+v)^{m+k}$$
which gives for the sum
$$[v^n] (1+v)^m
\sum_{k=0}^{n-1} \left\langle {n\atop k}\right\rangle
(1+v)^k.$$
Substitute the  generating function of the Eulerian  numbers into this
to get
$$[v^n] (1+v)^m
n! [z^n] \frac{v}{1+v-\exp(vz)}
\\ = [v^n] (1+v)^m
n! [z^n]
\frac{1}{1-(\exp(vz)-1)/v}.$$
Observe that $(\exp(vz)-1)/v$ expanded  about zero with respect to $z$
starts at $z.$ Since we  are extracting the $[z^n]$ coefficient we may
write
$$[v^n] (1+v)^m n! [z^n]
\sum_{p=0}^n \frac{(\exp(vz)-1)^p}{v^p}.$$
Now the coefficient extraction from the term in $z$ is
$$n! [z^n] (\exp(vz)-1)^p =
v^n  \times n! \times \frac{1}{v^n} [z^n] (\exp(vz)-1)^p
\\ = v^n \times n! [z^n] (\exp(z)-1)^p
= v^n \times p! \times {n\brace p}.$$
Returning to the main sum we thus obtain
$$[v^n] (1+v)^m
\sum_{p=0}^n v^{n-p} \times p! \times {n\brace p}
\\ = \sum_{p=0}^n [v^p] (1+v)^m \times p! \times {n\brace p}
= \sum_{p=0}^n {m\choose p} \times p! \times {n\brace p}.$$
To conclude observe that
$$m^n = n! [z^n] \exp(mz)
= n! [z^n] \sum_{p=0}^m {m\choose p} (\exp(z)-1)^p
\\ = \sum_{p=0}^m {m\choose p} \times p! \times {n\brace p}.$$
Since  the  effective  upper  limit  is always  $\min(n,m)$  which  is
symmetric we may use  either $m$ or $n$ as the upper  limit and we are
done.
Alternatively when $n \lt m$ we may change the upper limit from $m$ to
$n$ because the Stirling number is zero for $n\lt p \le m.$ If $n \gt
m$ we may also set the upper limit to $n$ because the binomial
coefficient is zero for $n\ge p \gt m.$

Remark.  This  last sum  may  also  be evaluated  combinatorially.
Count all functions from $[1,n]$ to  $[1,m].$ To do this first fix the
number  $p$ of different  values that  will occur,  next choose  a set
partition of $n$ into $p$ sets  the constituents of each of which will
be mapped  to the same value,  and finally choose the  $p$ values from
$m$ to use and some ordering thereof, giving the result
$$m^n =
\sum_{p=0}^n {n\brace p} \times {m\choose p} \times p!.$$
