It can be proven easily using geometric summation. Observe that $$ \sum_{k=0}^{r-1} \exp\left(i \frac{2\pi k m}{n} \right) = \sum_{k=0}^{r-1} \exp(2\pi i m/n)^k = \frac{1-\exp(2\pi i m/n)^r}{1-\exp(2\pi i m/n)} $$ The sum is $0$ if and only if $\exp(2\pi i m/n)^r = 1$, which corresponds to $n|rm$.