Q:

Is there a closed form solution for $\frac {d^n}{dx^n} \frac {1}{1-ae^x}$?

I am trying to find a closed form solution for $$\frac {d^n}{dx^n} \frac {1}{1-ae^x}$$
I have tried recognizing a pattern by computing specific cases but so far no luck.

I do know that the function is a composition of $x^{-1}$ and $1-ae^x$. The $k$th derivative of the latter is the same for all $k\ge 1$ and there is a closed form of the $k$th derivative of the former. Yet I am still not finding any answer.

However, when computing the derivatives manually I can see that in the numerator there is always a monic polynomial in $ae^x$ with constant term equal to one, all multiplied by $-ae^x$ and the denominator is always raised to the $n+1$ for the $n$th derivative.

I tried arranging the coefficients of the polynomial in the numerator in a triangle like that of Pascal's but I only recognized a pattern for the coefficients of the first and last two terms.

I tried Faa di Bruno's Formula but frankly it is over my head. Is there a clever solution or even a solution at all?

A:

By explicitly computing the first few derivatives it looks like the $n$'th derivative of $\frac{1}{1-ae^x}$ is on the form
$$\color{red}{\frac{d^n}{dx^n}\left(\frac{1}{1-ae^x}\right) = \frac{\sum_{k=1}^n f(n,k) a^ke^{kx}}{(1-ae^x)^{n+1}}}$$
where $f(n,k)$ are constants. This functional form can formally be proven by induction. By taking the derivative of the expression above and equating it to the same formula with $n\to n+1$ we get 
$$\frac{d}{dx}\left(\frac{\sum_{k=1}^n f(n,k) a^ke^{kx}}{(1-ae^x)^{n+1}}\right) = \frac{\sum_{k=1}^{n+1} f(n+1,k) a^ke^{kx}}{(1-ae^x)^{n+2}} $$
which when expanded gives us a recursion formula for $f(n,k)$:
$$f(n+1,k) = k f(n,k) + (n+2-k) f(n,k-1)~~~\text{for}~~~k=2,3,\ldots,n$$
with $f(n+1,1) = f(n+1,n+1) = 1$. This is the known recursion formula for the (shifted) Eulerian numbers $f(n,k) = A(n,k-1) \equiv \left\langle {n \atop k-1} \right\rangle$.
For more information about the Eulerian numbers see e.g. MathWorld : Euler's number triangle, Wiki : Eulerian Numbers, OEIS  : Eulerian Numbers. The first few values are (each row gives $f(n,k)$ for $k=1,2,\ldots,n$):
$$\matrix{1 & n=1\\1~~~1 & n=2\\1~~~4~~~1 & n=3\\1~~~11~~~11~~~1 & n=4\\1~~~26~~~66~~~26~~~1 & n=5\\1~~~57~~~302~~~302~~~57~~~1 & n=6}$$

