Classifying singularities of a complex function

[TheCanadian]

Homework Statement

[/B]
Find and classify the isolated singularities of the following:

$$ f(z) = \frac {1}{e^z - 1}$$

Homework Equations

The Attempt at a Solution

I have the solution for the positions of the singularities, which is: $z = 2n\pi i$ (for $n = 0, \pm 1, \pm 2, ...$) and this makes sense. But the solution also says these are simple poles. This part I do not quite understand. For example, if I take the derivative of $f(z)$, then $f'(z) = -(e^z -1)^{-2} e^z$ where $f'(2n\pi i)$ is also $\infty$. Perhaps my understanding of simple poles is wrong, but I don't see any singularities of order 1 when considering the Laurent series of $f(z)$.

Any ideas on why $z = 2n\pi i$ is considered a simple pole here?

 

[fresh_42]

If $z_0=1$ is a pole of first order of $f(z)$, what can be said about $(z-z_0)\cdot f(z)$?

 

[TheCanadian]

If $z_0=1$ is a pole of first order of $f(z)$, what can be said about $(z-z_0)\cdot f(z)$?

Ahh that makes sense. When evaluating $(z-z_0)\cdot f(z)$ at $z = z_0$, the value is finite. I incorrectly took the derivative above and treated this like a zero of order 1 instead of a pole. Thank you for the help.