<s>[INST] 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? 
 [/INST] 
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}

</s>