$\sum_{j=0}^{n-1}z_j^k=\begin{cases} 0, & \text{if $1\leq k \leq n-1$ } \\ n, & \text{if $k=n$ } \end{cases}$ Show that $\sum_{j=0}^{n-1}z_j^k=\begin{cases}
0,  & \text{if $1\leq k \leq n-1$ } \\
n, & \text{if $k=n$ }
\end{cases}$,  where $z_0,...,z_{n-1}$ are the $n$-th roots of unity.
For $k=n$ it is trivial as all the numbers will be $1$. Hence we get $1+\cdots +1$ ($n$ times). But is there any quick process for the rest?
 A: If $z_0$ is a primitve $n$-th root of unity, then $1, z_0^2,\ldots z_0^{n-1}$ are the remaining roots of unity.
So,
$$\sum_{j=0}^{n-1}z_j^k=\sum_{j=0}^{n-1}z_0^{kj}.$$
If $k=n$ this sum becomes
$$\sum_{j=0}^{n-1}1=n.$$
If $1\le k\le n$, then $z_0^k$ is an $n$-th root of unity, and so the sum vanishes as in Olivier Ochoa's answer.
A: Hint. You may use the identity

$$
1+z+z^2+\cdots+z^{n-1}=\frac{1-z^n}{1-z},\qquad z\neq 1, \tag1
$$ 

with $z:=z_j$ such that $z_j^n=1$ and $z_j \neq1$.
You may prove $(1)$ by evaluating
$$
(1-z)\left(1+z+z^2+\cdots+z^{n-1}\right).
$$ The case $z_j^n=1$ is direct.
A: Let $\omega = e^{\frac{2\pi}{n}}$. Then for $i = 0,1,\ldots,n-1$ we have $z_i = \omega^i$.
Let's just look at the case $k \neq n$.
If $\gcd(k,n) = 1$ then it's easy because raising every $n$-th root of unity to the $k$-th power is just permuting them so the sum remains the same as in the case of $k = 1$ (which is $0$).
If $\gcd(k,n) = d >1$ then raising every $n$-th root of unity to the $k$-th power creates $d$ identical sums 
$$(1+\omega^d+\ldots+\omega^{n/d-1})+(1+\omega^d+\ldots+\omega^{n/d-1})+\ldots+(1+\omega^d+\ldots+\omega^{n/d-1})$$
You'll find that each sum consists of a full set of the $n/d$-th roots of unity, and so is equal to $0$.
