Order of a zero - $\frac{1}{e^z-1}$ - Equivalence between two definitions I had to classify the singularity (removable, pole and essential) of $\displaystyle \frac{1}{e^z-1}$. 
I know that $e^z-1=0 \iff e^z=1 \iff z = 2\pi k i = z_k$ for each $k \in \mathbb{Z}$. In using what I found, the solution of the book explain that as the derivative of $e^z$ doesn't vanish at $z_k$, so each $z_k$ are a simple pole, i.e. $\displaystyle \lim_{z\to i2n\pi}\frac{z-i2n\pi}{e^z-e^{i2n\pi}}=\left. \frac{1}{\frac {de^z}{dz}}\right|_{z=i2n\pi}=1$.
According to Classify the singularity - $\frac{1}{e^z-1}$, Henry. W explains to me that if $\displaystyle \lim_{z→z_0} \frac{f(z)}{g(z)}=1$ (here $1$ is the order of the singularity; in general, It must be different from $0$ and $\infty$) (i.e. $f \sim g$) while $f(z_0)=0=g(z_0)$, the orders of their zeroes are the same. By the definition of zero, $f−g$ has the zero of same order at $z_0$. 
Here is the definition I know : Let $D \subset \mathbb{C}$ and a holomorphic function $f : D \to \mathbb{C}$. The function admits a zero of order $m \geq 1$ at $z_0 \in D$ if the Taylor series is on the form $\displaystyle f(z) = \sum_{k=m}^{\infty} a_k(z-z_0)^k$ with $a_m \not= 0.$ 
Is anyone could explain to me in details the difference between these two definition in the context of the problem?
Thanks!

If $f(z),g(z)$ both have zeroes of order $m$ at $z_0$, there exists holomorphic functions $\phi,\psi$ in a neighbourhood of $z_0$ such that $\phi(z_0) \neq 0$, $\psi(z_0) \neq 0 $ and $f(z) = (z-z_0)^m \phi(z), g(z) = (z-z_0)^m \psi(z)$, and thus the limit
$$ \lim_{z \to z_0} \frac{f(z)}{g(z)} = \lim_{z \to z_0} \frac{\phi(z) (z-z_0)^m}{\psi(z) (z-z_0)^m} = \frac{\phi(z_0)}{\psi(z_0)} \in \mathbb{C} \setminus \{0\}
$$
If the orders of their zeroes are different, then the above either $\to 0$(numer. order greater than denom.order) or grows unbounded(denom. order greater than numer.order)
