Let $\omega=e^{\frac{2\pi i}{n}}$.
Prove that $\Pi_{k=1}^{n-1} (1-\omega^k)=n.$
So far, I've tried brute-forcing it by expanding out the product, but it ended up getting too messy--and now I'm clueless as to how to proceed. I know the answer has to be fairly straightforward, but it's pretty late at night. Any suggestions would be appreciated.
$w^k$ for $k\in\text{{1,2,...n}}$ are roots of $z^n=1$
or, $z^n-1 \equiv(z-1)(z-w)(z-w^2)...(z-w^{\text{n-1}})$
also, $$(z-w)(z-w^2)...(z-w^{\text{n-1}})=\frac{z^n-1}{z-1}=(z^{n-1}+z^{n-2}+...+1)$$
plugging $z=1$
$(1-w)(1-w^2)...(1-w^{\text{n-1}})=(1+1...+1)=n$