Some identity related to Euler's identity

Question

Let $n\geq 2$, and $z=e^{\frac{2\pi i}{n}}$. Then, for $1\leq m\leq n-1$ we have the identity: $$ \sum_{j=0}^{n-1}z^{jm}=0. $$ Considering a proof of the identity, we can use an idea about periodicity from the unit circle. I am wondering if there is another way to prove it.

Answer 1

Note that the sum is a geometric sum; we get $$(1-z^m)\sum_{j=0}^{n-1}z^{jm}=\sum_{j=0}^{n-1}z^{jm}-\sum_{j=1}^nz^{jm} \stackrel{\ast}{=}z^0-z^n=0,$$ where $\ast$ holds because most terms of the two sums cancel out. Clearly $1-z^m\neq0$ so then the sum itself must equal $0$.