Does this sequence of polynomials have a name?

Question

I'm very interested in the function $$f : (0,\infty) \rightarrow (0,\infty)$$ $$x \mapsto - \log(1-e^{-x}).$$

When I use Wolfram alpha to compute the $n$th derivatives of $f$, I find that there exists a sequence of polynomials $P_1,P_2,\ldots$ such that for $n \geq 1$ we have $$f^{(n)}(x) = \frac{e^x}{(1-e^x)^n}P_n(e^x).$$

For example:

$$\frac{d^6}{dx^6}(-\log(1 - e^{-x})) = \frac{e^x}{(1-e^x)^6} (1+26 e^x + 66 e^{2 x} + 26 e^{3 x} + e^{4 x})$$

Question. Is there a name for this sequence of polynomials?

If these polynomials don't have a name, I'd also be satisfied with a name for the variant on Pascal's triangle whose entries are the coefficients of these polynomials.

That looks like an Eulerian polynomial: https://en.wikipedia.org/wiki/Eulerian_number — Angina Seng, Oct 03 '18 at 15:19
I will conjecture that we can describe the polynomials $P_n$ with a three term recurrence relation. Have you tried to work something like that out? — hardmath, Oct 03 '18 at 15:21

score 4 · Accepted Answer · answered Oct 03 '18 at 16:18

Lord Shark the Unknown is spot on.

WA tells us that $$ \frac{d}{dx}\frac{e^x}{(1-e^x)^n} P(e^x) = \frac{e^x}{(1-e^x)^{n+1}} Q(e^x) $$ where $$ Q(t)=(t-t^2)P'(t)+((n-1)t+1)P(t) $$ Therefore, $$ P_{n+1}(t)=(t-t^2)P_n'(t)+((n-1)t+1)P_n(t) $$ Since $P_2=A_1$, we have $P_n=A_{n-1}$, where $A_n$ is the $n$-th Eulerian polynomial.

Does this sequence of polynomials have a name?

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