Understanding how nth root can equal 1
Solution 1:
You wish to prove $$\prod_{k=1}^{n-1}(1-\omega^k)=n.$$
Since the $z=\omega^k$ (for $0\leq k<n$) are precisely the roots of $z^n-1$, we obtain $$z^n-1=\prod_{k=0}^{n-1}(z-\omega^k).$$
Dividing both sides by $z-1$ yields $$z^{n-1}+\cdots +z+1=\prod_{k=1}^{n-1}(z-\omega^k).$$
Substituting $z=1$ gives the result.