Power series of $x/(1-ae^{-x})$.

I am looking for a power series expansion for the function $x(1-ae^{-x})^{-1}$ (perhaps for $0<a<1$).

Using the Bernoulli numbers, I can write \begin{align*} \frac{x}{1-ae^{-x}} &= \frac{x}{1-e^{-(x -\log(a) )}}\\ &= \sum_{n\geq 0} \frac{(-1)^n B_n (x-\log(a))^n }{n!} \\ &= \sum_{n\geq 0} \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{B_n (\log(a))^{n-k} x^k }{n!} \\ &= \sum_{n\geq 0} (-1)^n \left( \sum_{m\geq n} \frac{\binom{m}{n} (\log(a))^{m-n} B_m}{m!} \right) x^n \end{align*} My question is, is there any neater power series expression for the function, in the sense that I don't have to evaluate another infinite sum for each coefficient of $x^n$? Or can we identify the number inside the large bracket of the above?

This is "almost" the generating function of the Eulerian polynomials: for $|a|<1$ we have \begin{align*}\frac1{1-ae^{-x}}&=\sum_{k=0}^\infty a^k\sum_{n=0}^\infty\frac{(-kx)^n}{n!}=\sum_{n=0}^\infty\frac{(-x)^n}{n!}\sum_{k=0}^\infty k^n a^k\\&=\frac1{1-a}\left(1+a\sum_{n=1}^\infty\frac{A_n(a)}{(1-a)^n}\frac{(-x)^n}{n!}\right).\end{align*} It's not hard to see that this actually holds for any $a\neq 1$. To get a "working" limit $a\to 1$ (that would lead to the Bernoulli numbers), multiply the above by $x-\log a$ (and not just by $x$).