Theorem. We have that $\displaystyle \underbrace{x\left(\dfrac{d}{dx}\left(\cdots x \left(\dfrac{d}{dx} \left( \dfrac{x}{1-x}\right)\right)\cdots\right)\right)}_{\text{$x \frac{d}{dx}$ $m$ times}}=\dfrac{1}{(1-x)^{m+1}}\sum_{k=0}^{m-1} A(m,k) x^{m-k}$ where $A(a,k)$ are the Eulerian numbers and $m$ is a nonnegative integer.

I am looking at the polylogarithm function $\operatorname{Li}_{-s}(z) = \displaystyle \sum_{k=1}^{\infty} \dfrac{z^k}{k^{-s}}$ and trying to find its formula for nonnegative integers $s$. Since it is easily verified that

$$\displaystyle \sum_{k=1}^{\infty}k^m x^k =\underbrace{x\left(\dfrac{d}{dx}\left(\cdots x \left(\dfrac{d}{dx} \left( \dfrac{x}{1-x}\right)\right)\cdots\right)\right)}_{\text{$x \frac{d}{dx}$ $m$ times}},$$ the result follows from this. I have noticed that the change of variables $x = e^t$ gives us $$\underbrace{x\left(\dfrac{d}{dx}\left(\cdots x \left(\dfrac{d}{dx} \left( \dfrac{x}{1-x}\right)\right)\cdots\right)\right)}_{\text{$x \frac{d}{dx}$ $m$ times}} = \dfrac{d^m}{dt^m}\left(\dfrac{1}{1-e^t} \right),$$ but I am wondering how to utilize this to prove the theorem.