If $1, \omega, \omega^2,..., \omega^{n-1}$ are the nth roots of unity then evaluate $(1-\omega)(1-\omega^2)...(1-\omega^{n-1})$
I think there might be walkthrough solution somewhere for this, and I did find one tho. But the thing is the answer is not the same Here is the link to the video I found
And here is my solution \begin{equation} z^n-1=(z-1)(z-\omega)...(z-\omega^{n-1})\\z^n-1=(z-1)(1+z+z^2+...+z^{n-1})\\\Rightarrow \prod_{k=1}^{n-1} (z-\omega^k)=\sum_{m=0}^{n-1} z^m\\z=1\Rightarrow \prod_{k=1}^{n-1} (1-\omega^k)=n \end{equation} Which the answer in the video say it equals to 0. Or perhaps $z$ can't not be equal 1? I'm very confused. Also in the video, why he picks $\omega^3$ why not $\omega$ or $\omega^2$?
