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.

  • 2
    $\begingroup$ Do you have any ideas on how to proceed? Have you tried anything? It is usually best not to copy a problem out of your problem list here... (One particular problem with your post is that you do not give any indication at all about what you know, and therefore it is very easy for someone to come and write an answer which is waaay over your head, and similar phenomena) $\endgroup$ Commented Feb 4, 2015 at 7:00
  • $\begingroup$ Sorry, I'll edit what I've tried into the post. $\endgroup$ Commented Feb 4, 2015 at 7:03
  • $\begingroup$ See math.stackexchange.com/a/806679/589. $\endgroup$
    – lhf
    Commented Feb 4, 2015 at 10:39

1 Answer 1


$w^k$ 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 $z=1$


  • $\begingroup$ Very well done indeed, plus one! $\endgroup$ Commented Feb 4, 2015 at 7:58
  • $\begingroup$ @RobertLewis thank u very much :) $\endgroup$
    – Shobhit
    Commented Feb 4, 2015 at 8:18
  • $\begingroup$ And you are more than welcome! Cheers! $\endgroup$ Commented Feb 4, 2015 at 8:26
  • $\begingroup$ Of course, the correct answer, I just think the "or" on second line is not thoroughly appreciated. $\endgroup$
    – orangeskid
    Commented Feb 4, 2015 at 8:28
  • $\begingroup$ @orangeskid edit it however u see fit :) $\endgroup$
    – Shobhit
    Commented Feb 4, 2015 at 8:32

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