Taylor series of $\frac{1}{1-\exp(z-1)}$ As part ofa voluntary homework assignment, I want to find the Taylor series around 0 for the complex valued function
\begin{equation}
f(z):=\frac{1}{1-\exp(z-1)}
\end{equation}
My approach so far is to express $\exp(z-1)$ and $\frac{1}{1-z}$ as power series. Since $\exp(z)= \sum \frac{z^k}{k!}$ and $\frac{1}{1-z} = \sum{z^k}$, this leads me to
\begin{equation}
\frac{1}{1-\exp(z-1)} = \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \left( \frac{(z-1)^m}{m!} \right)^k = \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \left( \frac{(z-1)^{mk}}{(m!)^k} \right)
\end{equation}
To obtain a power series around 0, I use the binomial series:
\begin{equation}
\sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \left( \frac{(z-1)^{mk}}{(m!)^k} \right) = \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \sum_{n=0}^{mk} \frac{1}{(m!)^k} \binom{mk}{n} z^n \cdot (-1)^{mk-n}
\end{equation}
However, I am not sure how to proceed at this point. Can the last expression be simplified?
Of course, one could find the solution by using derivatives, but I should find the coefficients of the Taylor series based on the geometric and exponential power series. Am I overlooking something?
 A: HINT: An easier place to start would be 
$$
\frac{1}{1-e^{z-1}} = \sum_{n=0}^{\infty}\frac{e^{nz}}{e^n}
$$
Then expand the exponential $e^{nz}$ in the variable $nz$, instead of raising a power series to an exponent.
A: Following  Himadri's suggestion:
\begin{equation}
f(z):=\frac{1}{1-e^{-1}\exp(z)} = \sum_{k=0}^{\infty} e^{-k}e^{zk} = \sum_{k=0}^{\infty} e^{-k} \sum_{m=0}^{\infty} \frac{(zk)^{m}}{m!}
\end{equation}
Now exchange the summation orders:
\begin{equation}
f(z) =\sum_{m=0}^{\infty} \frac{z^m}{m!}\; \sum_{k=0}^{\infty} e^{-k}  k^m =\sum_{m=0}^{\infty} \frac{z^m}{m!}\; c_m
\end{equation}
So the difficulty arises in evaluating the coefficients 
$c_m = \sum_{k=0}^{\infty} e^{-k} k^m$. Write
$$
\sum_{k=0}^{\infty} e^{-k} k^m = \lim_{a\to -1}\sum_{k=0}^{\infty} e^{ak} k^m \\
= \lim_{a\to -1}\sum_{k=0}^{\infty}  \frac{\partial^m}{\partial a^m}e^{ ak} =
\\
= \lim_{a\to -1} \frac{\partial^m}{\partial a^m}\sum_{k=0}^{\infty} e^{ak} =
\\
= \lim_{a\to -1} \frac{\partial^m}{\partial a^m}\frac{1}{1-e^{a}}
$$
and proceed from there. A few coefficients are (note they do not include the $1/m!$): 
$$
\begin{align}
c_0 &= \frac{e}{e-1} &\\
c_1 &= \frac{e}{(e-1)^2} &\\
c_2 &= \frac{e(1+e)}{(e-1)^3} &\\
c_3 &= \frac{e(1 + 4e + e^2)}{(e-1)^4} &\\
c_4 &= \frac{e (1 + 11 e + 11 e^2 + e^3)}{(e-1)^5} &
\end{align}
$$
They can be written as a finite sum $c_m = e\sum_{j=0}^{m}{m\brace j}\frac{j!}{\left(e-1\right)^{j+1}}$, see Markus Scheuer's answer.
A: 
I think it is more convenient to start with
  \begin{align*}
f(z)=\frac{1}{1-e^{z-1}}&=\sum_{n=0}^\infty e^{(z-1)n}
\end{align*}
  and calculate the Taylor series via
  \begin{align*}
f(z)=\sum_{n=0}^\infty f^{(n)}(0)\frac{z^n}{n!}
\end{align*}
We obtain
  \begin{align*}
\color{blue}{\left.\left(\frac{d^k}{dz^k}f(z)\right)\right|_{z=0}}&=\left.\sum_{n=0}^\infty n^ke^{(z-1)n}\right|_{z=0}\tag{1}\\
&=\sum_{n=0}^\infty n^k\left(\frac{1}{e}\right)^n\tag{2}\\
&=\sum_{n=0}^\infty \sum_{j=0}^k{k\brace j}n^{\underline{j}}\left(\frac{1}{e}\right)^n\tag{3}\\
&=\sum_{j=0}^k{k\brace j}\left(\frac{1}{e}\right)^j\sum_{n=0}^\infty n^{\underline{j}}\left(\frac{1}{e}\right)^{n-j}\tag{4}\\
&=\sum_{j=0}^k{k\brace j}\left(\frac{1}{e}\right)^j\left.\left(\frac{d^j}{dz^j}\frac{1}{1-z}\right)\right|_{z=\frac{1}{e}}\tag{5}\\
&=\sum_{j=0}^k{k\brace j}\left(\frac{1}{e}\right)^j\frac{j!}{\left(1-\frac{1}{e}\right)^{j+1}}\tag{6}\\
&\,\,\color{blue}{=e\sum_{j=0}^{k}{k\brace j}\frac{j!}{\left(e-1\right)^{j+1}}}\\
\end{align*}

Comment:


*

*In (1) we differentiate $k$ times.

*In (2) we evaluate the expression at $z=0$.

*In (3) we represent $n^k$  with the help of the Stirling numbers of the second kind as sum of falling factorials $n^{\underline{j}}=n(n-1)\cdots(n-j+1)$.

*In (4)  we exchange the sums.

*In (5) we use the geometric series expansion.

*In (6) we differentiate $j$ times  and evaluate at $z=\frac{1}{e}$.

We conclude
  \begin{align*}
\color{blue}{\frac{1}{1-e^{z-1}}=e\sum_{n=0}^\infty \left(\sum_{j=0}^{n}{n\brace j}\frac{j!}{\left(e-1\right)^{j+1}}\right)\frac{z^n}{n!}}
\end{align*}

