Finding general coefficients for $\frac{\partial^n}{\partial x^n} \left( \frac{1}{1 - e^{-x}} \right)$ I want to compute the following $n-th$ differentiation
\begin{align}
\frac{\partial^n}{\partial x^n} \left( \frac{1}{1 - e^{-x}} \right)
\end{align}
With the some help of computation tools i have
\begin{align}
\frac{\partial^n}{\partial x^n} \left( \frac{1}{1 - e^{-x}} \right) 
= (-1)^n \sum_{k=1}^n a(k,n) \frac{e^-kx}{(1-e^{-x})^{k+1}}
\end{align}
Here what i want to do is find exact expression for $a(k,n)$
Here i state some examples 
\begin{align}
n=1 \qquad : \qquad -\frac{e^{-x}}{\left(1-e^{-x}\right)^2} \\
n=2 \qquad : \qquad \frac{2 e^{-2 x}}{\left(1-e^{-x}\right)^3}+\frac{e^{-x}}{\left(1-e^{-x}\right)^2} \\
n=3 \qquad : \qquad  -\frac{6 e^{-2 x}}{\left(1-e^{-x}\right)^3}-\frac{6 e^{-3 x}}{\left(1-e^{-x}\right)^4}-\frac{e^{-x}}{\left(1-e^{-x}\right)^2} \\
n=4 \qquad : \qquad \frac{14 e^{-2 x}}{\left(1-e^{-x}\right)^3}+\frac{36 e^{-3 x}}{\left(1-e^{-x}\right)^4}+\frac{24 e^{-4 x}}{\left(1-e^{-x}\right)^5}+\frac{e^{-x}}{\left(1-e^{-x}\right)^2} \\
n=5 \qquad : \qquad   -\frac{30 e^{-2 x}}{\left(1-e^{-x}\right)^3}-\frac{150 e^{-3 x}}{\left(1-e^{-x}\right)^4}-\frac{240 e^{-4 x}}{\left(1-e^{-x}\right)^5}-\frac{120 e^{-5 x}}{\left(1-e^{-x}\right)^6}-\frac{e^{-x}}{\left(1-e^{-x}\right)^2} 
\end{align}
and so on. 
I am trying to construct the coefficients but got stuck. 
can you have any ideas? 
 A: Not an answer, but a recurrence and generating function.
Let $f(z)=\frac{1}{1-z}$. Let $g(x)=f(e^{x})$. Your function is $g(-x)$, but it is easier to avoid the complication of the $-x$ initially.
$$\begin{align}g'(x)&=-f'(e^{x})e^{x}\\
g''(x)&=f''(e^{x})e^{{2x}}+f'(e^{x})e^{x}\\
...
\end{align}$$
In general, if $$g^{(n)}(x)=\sum_{k=1}^{n} b_{n,k}f^{(k)}(e^x)e^{kx}$$
Then $$g^{(n+1)}(x)=\sum_{k=1}^{n}kb_{n,k}f^{(k)}(e^x)e^{kx}+\sum_{k=1}^{n}b_{n,k}f^{(k+1)}(e^x)e^{(k+1)x}$$
So $b_{n+1,k}=kb_{n,k}+b_{n,k-1}$, where $b_{0,0}=1$ and $b_{0,k}=0$ if $k\neq 0$.
Now, for $n>0$, $f^{(k)}(z)=\frac{k!}{(1-z)^{k+1}}$. So $$a_{n,k}=(-1)^{n}k!b_{n,k}.$$
Defining $F(u,v)=\sum_{n,k=0}^{\infty} b_{n,k}u^kv^n$, the recurrence will give the differential equation:
$$F(u,v)=1+uv\left(\frac{\partial F}{\partial u}+F(u,v)\right)$$
or:
$$\frac{\partial F}{\partial u} = \frac{(1-uv)F(u,v)-1}{uv}$$
A: 
We can write
  \begin{align*}
\frac{d^n}{dx^n}\left(\frac{1}{1-e^{-x}}\right)=(-1)^n\sum_{k=1}^n{n\brace k}k!\frac{e^{-kx}}{\left(1-e^{-kx}\right)^{k+1}}
\end{align*}

with the  numbers ${n\brace k}$ denoting the Stirling numbers of the second kind.

The coefficients $a(k,n)={n\brace k}k!$ are archived as OEIS/A019538 and start with
\begin{array}{c|rrrrr}
{n\brace k}k!&1&2&3&4&5\\
\hline
1&1\\
2&1&2\\
3&1&6&6\\
4&1&14&36&24\\
5&1&30&150&240&120\\
\end{array}

A: Let $\textbf{v}_{n}=\frac{1}{(1-e^{-x})^{n}}$. We have $D_{x} \cdot \textbf{v}_{n}=\frac{-n\cdot e^{-x}}{(1-e^{-x})^{n+1}}=n\cdot(\textbf{v}_{n}-\textbf{v}_{n+1})$. You can set up a matrix to represent $D_{x}$ to find your coefficients by letting $\textbf{v}_{n}$ be the nth basis vector of the standard basis, and the powers of said matrix will tell you the coefficients.
