Question: $z^{n}=1$

to prove that
$1+z+z^{2}+z^{3}+...+z^{n-1}=0$

i think I have to start with

$1+z^{2}+z^{3}+...+z^{N}=\frac{1-z^{N+1}}{1-z}$where $N=n-1$

I don't know where to go from there

Answer: 
If $z^n=1$ then $1-z^n=0$. Now if $z \ne 1$, then (as you wrote)
$$
1+z^{2}+z^{3}+...+z^{n-1}=\frac{1-z^{n}}{1-z}
$$
and the RHS is zero, so the LHS is zero.
If $Z^n = 1$

$Z$ is the nth root of unity, since $(1)^{any number}$ is always $1$, Therefore $z =1$ is one of the solution

Then $Z^n-1 = 0$ is always a multiple of $Z-1$

Use long division to divide them

$\frac{Z^n-1}{Z-1}$ this gives us a polynomial with all its coefficients equal to one $(1)$

$\frac{Z^n-1}{Z-1} = 1+Z+Z^2+Z^3+.........+Z^{n-1}$