A sum of powers of primitive roots of unity

Question

For the primitive roots of unity $\omega_n = e^{i2\pi/n}$ I want to prove that $$\sum_{k=0}^{n-1} \omega_n^{lk} = 0$$ if $n$ doesn't divide $l$.

I have already proven the well-known result $$\sum_{k=0}^{n-1} \omega_n^{k} = 0$$ so I only need to show that if I raise the $\omega_n$s to the power of $l$, I get every power of $w_n$ exactly once. This is a simple algebraic statement but I don't see how to prove it.

Answer 1

Say that sum is $s$. Show that $\omega_n^l s=s$ and that $\omega_n^l\ne 1$.

Answer 2

Hint: Show that $(\omega_n^{\ell 1},\ldots,\omega_n^{\ell (n-1)})$ gives a permutation of $(\omega_n^{1},\ldots,\omega_n^{n-1})$ when $\ell$ and $n$ are coprime.