Type of singularity in $\frac{1}{1-e^z}$

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.

$$ \frac1{1-e^z}=\sum_{n\ge0}e^{zn}\tag1 $$ 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 $z=0$.