Number of zeros of $1-e^{z^k}$ in the complex plane.

Question

Let $k$ be a positive integer and consider the function $f(z) = 1-e^{z^k}$ in the complex plane. Why is it true that the number of zeros of $f$ in a disk of radius $r$ is about $\pi^{-1}kr^k?$ I tried using Rouche's theorem, but the integral looked quite messy. Is there a better way to see this?

Answer 1

Hint

$$1-e^{z^k}=0\iff e^{z^k}=1\iff z^{k}=2m\pi,\quad m\in\mathbb Z$$