Does $\displaystyle \underbrace{x\left(\dfrac{d}{dx}\left(\cdots x \left(\dfrac{d}{dx} \left( \dfrac{x}{1-x}\right)\right)\cdots\right)\right)}_{\text{$x \frac{d}{dx}$ $m$ times}}$ have a closed form?
I am curious about this because of this question. Testing this out for a few values of $m$ I get:
$m = 1: \dfrac{x}{(x-1)^2}$
$m = 2: \dfrac{-x^2-x}{(x-1)^3}$
$m = 3: \dfrac{x^3+4x^2+x}{(x-1)^4}$
$m = 4: \dfrac{-x^4-11x^3-11x^2-x}{(x-1)^5}$
$m = 5: \dfrac{x^5+26x^4+66x^3+26x^2+x}{(x-1)^6}$
$\vdots$
It is easy to see that the closed form will have a $(-1)^{n+1}$ in it and seeing how some of the terms look symmetric, it will probably have something like $(x+1)^n$, but I don't see a way to find the closed form.
Edit: I want to prove that $$\operatorname{Li}_{-a}(z)= \displaystyle \sum_{k = 1}^{\infty} k^m x^k=\frac 1{(1-z)^{m+1}}\sum_{k=0}^{m-1} E(m,k)\, z^{m-k}$$ with $E(a,k)$ the Eulerian numbers.
Note : $$\operatorname{Li}_{-m}(z)= \displaystyle \sum_{k = 1}^{\infty} k^m z^k=\frac 1{(1-z)^{m+1}}\sum_{k=0}^{m-1} E(m,k)\, z^{m-k}$$ with $E(m,k)$ the Eulerian numbers.
This relationship is wellknown. For example, see eqs.(4) and (5) in : http://mathworld.wolfram.com/EulerianNumber.html
With the correspondance of symbols : $z=r$ , $m=n$ , $k=i$ , $E(m,k)=\Big\langle \begin{array}{c} n \\ i \end{array} \Big\rangle$
Answer of the question about closed form :
The change of variable $x=e^t$ simplifies a lot the equation :
$$\displaystyle \underbrace{x\left(\dfrac{d}{dx}\left(\cdots x \left(\dfrac{d}{dx} \left( \dfrac{x}{1-x}\right)\right)\cdots\right)\right)}_{\text{$x \frac{d}{dx}$ $m$ times}\; ,\; m>0} = \frac{d^m}{dt^m}\left(\frac{1}{1-e^t} \right)$$
I don't think that a simpler expression exists for the $m$-th derivative (other than the polylogarithm in this particular case, of course).