Originally Posted by

**Deveno** i have a totally different suggestion, consider instead the expression:

$\displaystyle \prod_{r=1}^{n-1} |\cos \left(\frac{2\pi r}{n}\right) + i \sin \left(\frac{2\pi r}{n}\right) - x| = \left|\prod_{r=1}^{n-1} (\cos \left(\frac{2\pi r}{n}\right) + i \sin \left(\frac{2\pi r}{n}\right) - x)\right|$

note that for each r, we have $\displaystyle \cos \left(\frac{2\pi r}{n}\right) + i \sin \left(\frac{2\pi r}{n}\right)$ is a root of $\displaystyle 1 - x^n$,

and we have every root except the root 1 represented in our product of factors, so:

$\displaystyle \prod_{r=1}^{n-1} (\cos \left(\frac{2\pi r}{n}\right) + i \sin \left(\frac{2\pi r}{n}\right) - x) = \frac{1-x^n}{1-x} = 1 + x + \dots + x^{n-1}$,

hence $\displaystyle \prod_{r=1}^{n-1} |\cos \left(\frac{2\pi r}{n}\right) + i \sin \left(\frac{2\pi r}{n}\right) - x| = |1 + x + \dots + x^{n-1}|$.

now just let x = 1, and voila!

(edit: by the way, the formula still makes sense for n = 2, and it is true for n = 2...it's just not very interesting, there it boils down to |(-1) - 1| = 2, which is fairly obvious).