Singularities of $1\over e^z - 1$

Note \(e^z = 1\) when \(z=z_k\da 2\pi k\) for \(k\in \ZZ\), and the claim is that these are all poles of order 1 of \(f(z)\). These are poles of some order: each \(z_k\) is a zero of \(1/f\) (since \(e^{z_k} - 1 = 0\)), so \(f\) has a singularity at \(z_k\); and the order of the pole is the smallest \(n\) for which \(\lim_{z\to z_k}(z-z_k)^n f(z)\) exists. Start by computing the first: \[ \lim_{z\to z_k}(z-z_k)f(z) = \lim_{z\to z_k} {z-z_k\over e^z - 1} \equalsbecause{\text{LH}} \lim_{z\to z_k} {1\over e^z} = e^{-z_k} = 1 .\]