Type of singularity in $\frac{1}{1-e^z}$ My complex analysis textbook uses the following definitions for removable singularity, essential singularity and for poles of a complex function:

If $a$ is an isolated singularity of $f$ and the numbers $c_n$ for $n \in \mathbb{Z}$ are the coefficients in the Laurent series of the function, $\sum_{n\in\mathbb{Z}} c_n(z-a)^n$, then:

*

*$a$ is a removable singularity if $c_n = 0$ for $n < 0$


*$a$ is a pole of order $m \in \mathbb{N}$ if $c_{-m} \neq 0$ and $c_n=0$ for $n < -m$


*$a$ is an essential singularity if $c_n \neq 0$ for an infinite number of negative values of $n$

Then, based on this definition I am supposed to classifies the singularity of the function $f(z)= \frac{1}{1-e^z}$ at the point $z = 0$
So, first I began by finding the Laurent series of this funcion:
$$\begin{align}
\frac{1}{1-e^z}&=\sum_{n \geq0}e^{zn}
\end{align}$$
Because we have that $e^z=\sum_{k \geq 0}\frac{1}{k!}z^k$. So if we let $z=zn$ then we get:
$$\begin{align}
\sum_{n \geq0}e^{zn} &= \sum_{n \geq0} \sum_{k \geq 0}\frac{n^k}{k!}z^k
\\
\\
&= \sum_{k \geq0}\  \underbrace{ \sum_{n \geq 0} \frac{n^k}{k!}}_{:=a_k}\ z^k
\\
\\
&= \sum_{k \geq0} a_k z^k
\end{align}$$
So, according to the definition my book gave, this is a removable singularity. But when I checked the solutions it said that $z=0$ is a pole of order $1$. So did I do something wrong and if so what did I do wrong or is the did the book author make a mistake?
 A: 1) An isolated singularity is removable in the sense that one can define limit of the function at that point where singularity is. In this case, $\lim_{z\to 0} \frac{1}{1-e^z}=\frac{1}{1-e^0}=\frac{1}{1-1} $  which you can't define.     
$ \\ $
2) We say a point $z_0$ is a pole if by multiplying some power of $(z-z_0)$ with the function $f(z)$ , you can kill the singularity in the sense that multiplying $f(z)$ with some power of $(z-z_0)$ you can define the value at $z_0$  of the resulting function and the least power of $(z-z_0)$, you need to do multiply is called the order of the pole. In this case, $z=0$ is an isolated singularity and $\lim_{z\to 0} \frac{z}{1-e^z}=-1$ . Hence, $0$ is a pole of order 1. 
A: $$
\frac1{1-e^z}=\sum_{n\ge0}e^{zn}\tag1
$$
is not the Laurent series for $\frac1{1-e^z}$. It is one series for $\frac1{1-e^z}$ that only converges for $\operatorname{Re}(z)\lt0$, so probably not good for use in finding the type of singularity of $\frac1{1-e^z}$ at $z=0$.
The Laurent series for $\frac1{1-e^z}$ at $z=0$ is a little messy to compute, but suppose we have the Laurent series at $z=0$
$$
\frac1{1-e^z}=\sum_{k=-n}^\infty a_kz^k\tag2
$$
The $n$ we want to find is the smallest $n$ so that $\lim\limits_{z\to0}\frac{z^n}{1-e^z}$ is finite.
If $\lim\limits_{z\to0}\frac1{1-e^z}$ were finite, then $n=0$. However, this limit is $\infty$.
Otherwise, if $\lim\limits_{z\to0}\frac z{1-e^z}$ were finite, then $n=1$. Using L'Hôpital, we get that
$$
\begin{align}
\lim_{z\to0}\frac z{1-e^z}
&=\lim_{z\to0}\frac1{-e^z}\\
&=-1\tag3
\end{align}
$$
Thus, $n=1$. So $\frac1{1-e^z}$ has a pole of order $1$ at $z=0$.
A: By your definition, $$a_0 = \sum_{n \geq 0} \frac{n^0}{0!} = \sum_{n \geq 0} 1 = \infty,$$ so $a_0$ can't be a coefficient of the Laurent series.
To show it's actually a pole of order 1, you could verify that $\lim_{z \rightarrow 0} \frac{z}{e^z - 1} < \infty$. 
