$\frac{1}{2\pi i} \int_\gamma z^m\frac{f'(z)}{f(z)}dz=\sum_k (z_k)^m$

Question

This question comes from Joseph Bak Complex Analysis.
Suppose f is analytic inside and on a regular closed curve γ and has no zeroes on γ . Show that if m is a positive integer then $ \frac{1}{2\pi i} \int_\gamma z^m \frac{f'(z)}{f(z)}dz=\sum_k (z_k)^m $, where the sum is taken over the zeroes of f inside $\gamma$.
I can't realy use the claim that it's the number of zeores of $z^m f$ , and I can't see how it follow from the residue thm. Any hint will be great!

Answer 1

It follows from the generalized argument principle that$$\frac{1}{2\pi i} \int_\gamma z^m\frac{f'(z)}{f(z)}dz=\sum_{z_k\in Z} (z_k)^m\eta(\gamma,z_k)-\sum_{w_k\in P} (w_k)^m\eta(\gamma,w_k),$$where $Z$ is the set of zeros of $f$ and $P$ is the set of its poles. But there are no poles here and all those winding numbers $\eta(\gamma,z_k)$ are equal to $1$.