Simple prove that product of the diagonals of a polygon = N

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.

Solution 1:

Here is an extremely simple proof but it requires working in the field $\mathbb{C}$.

For any natural $n > 0$:

  Let $w = e^{i\frac{2\pi}{n}}$

  $\prod_{k=1}^n (z-w^k) = z^n-1$ because both are monic polynomials of degree $n$ with the same $n$ roots

  Thus $\prod_{k=1}^{n-1} (z-w^k) = \lim_{t \to z} \frac{t^n-1}{t-1} = \sum_{k=0}^{n-1} z^k$

  In particular $\prod_{k=1}^{n-1} (1-w^k) = \sum_{k=0}^{n-1} 1^k = n$

  Therefore the product of the diagonals desired is $|n| = n$

Solution 2:

Consider the complex solutions $z_0, \ldots , z_{n-1}$ to $(z+1)^n-1 = 0$. These solutions are evenly spaced on the circle centered at $z = -1$ with radius $1$. Thus, they form a regular $n$-gon.

Since $0 = (z+1)^n-1 = z^n + \cdots + nz+1-1 = z(z^{n-1}+\cdots +n)$, we know that $z_0 = 0$ is one solution, and the product of the other $n-1$ solutions is $z_1 \cdots z_{n-1} = (-1)^{n-1}n$.

Then, the product of the distances from $z_0$ to each of the $z_i$'s is $|z_1-z_0| \cdots |z_{n-1}-z_0| = |z_1| \cdots |z_{n-1}| = |z_1 \cdots z_{n-1}| = |(-1)^{n-1}n| = n$, as desired.