TITLE: Simple prove that product of the diagonals of a polygon = N
QUESTION [21 upvotes]: There is a beautiful fact:

If you take a regular N-sided polygon, which is inscribed in the unit circle and find the product of all its diagonals (including two sides) carried out from one corner you will get N exactly:
$A_1A_2\cdot A_1A_3\cdot ...\cdot A_1A_N = N$

For example, for a square we have $\sqrt{2}\cdot 2\cdot \sqrt{2} = 4$.

I know there is some prove, which is based on complex numbers. But the result is so simple that I wonder is there much more simple prove, which you can explain to a school boy easily?
P.S. Please, use spoiler tag >! in your answers.

REPLY [4 votes]: Here is my attempt at an elementary proof. After hours of staring, I believe (at minimum) a few trigonometric identities would be required in any elementary demonstration. 
Given our circle with $n$ nodes going around the circle, it is clear that each edge of interest, $a_k$, $1\le k\le n-1$, is part of an isosceles triangle with node at the center, sides, $1,1,a_k$, and inner angle $(2\pi)/n$. Thus, using law of cosines and a product to some formula, we gain, 
$$a_k^2=1^2+1^2-2\cos\left(\frac{2\pi k}{n}\right)=4\sin\left(\frac{k\pi}{n}\right)\sin\left(\frac{(n-k)\pi}{n}\right).$$ We now have, using this (formula 24), 
$$a_1^2\cdots a_{n-1}^2 = \prod_{k=1}^{n-1}\left(4 \sin\left(\frac{k\pi}{n}\right)\sin\left(\frac{(n-k)\pi}{n}\right)\right) =\\ 4^{n-1} \left(\prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right)\right)^2 = 4^{n-1}\left(\frac{n}{2^{n-1}}\right)^2 = n^2.$$ Hence, our theorem holds, $$a_1\cdots a_{n-1}=n.$$