Product of $ |z^k - 1| $ Problem: Prove the following identity about the product involving the nth roots of unity:
$$
\prod_{k=1}^{N-1}|z^k-1| = N
$$
where $ z^k $ is the primitive nth root of unity.

Attempt:
$$
\begin{align}
\prod_{k=1}^{N-1}|z^k-1| &= \prod_{k=1}^{N-1}\left|(\cos(\frac{2\pi k}{N})-1)+i\sin(\frac{2\pi k}{N})\right| \\
&=\prod_{k=1}^{N-1}\sqrt{\cos^2(\frac{2\pi k}{N})-2\cos(\frac{2\pi k}{N})+1+\sin^2(\frac{2\pi k}{N})} \\
&=\prod_{k=1}^{N-1}\sqrt{2-2\cos(\frac{2\pi k}{N})} \\
&=\prod_{k=1}^{N-1}2\sqrt{\frac{1}{2}-\frac{1}{2}\cos(\frac{2\pi k}{N}))} \\
&=2^{N-1}\prod_{k=1}^{N-1}\sin(\frac{k\pi}{N})
\end{align}
$$
I found on Wikipedia that there is an identity for the last product: $ \prod_{k=1}^{N-1}\sin(\frac{k\pi}{N}) = N/2^{N-1} $. However I do not know how to prove it.
Could someone help me prove the last identity or perhaps suggest a different approach to the problem?

Aha, I just solved it:
First consider the polynomial $$ \prod_{k=0}^{N-1}(x-z^k) $$
The roots of the polynomial are the nth roots of unity, which are precisely the roots of the polynomial $ x^n-1 $ and so the two are equal.
Dividing both sides by $ x-1 $, we get
$$ \prod_{k=1}^{N-1}(x-z^k) = 1+x+\dots+x^{N-1} $$
Substituting $ x=1 $, we get that the product equals $ N $.
The product of the magnitudes is simply the magnitude of the product, so we get the desired result $$ \prod_{k=1}^{N-1}|1-z^k| = N $$
