I want to derive this formula
$$r_1^2.r_2^2.r_3^2.....r_n^2=r^{2n}+c^{2n}-2r^nc^n\cos(n\theta)$$
where $r_1, r_2, r_3....r_n$ are distances of a Point $P(r,\theta)$ from equally spaced n points, say $A_1, A_2, A_3....A_n$ on a circle of radius $c$ with $O$ as the centre of the circle as well as the origin
As the circle is divided into n equal part, we can say $$\angle A_1OA_2=\angle A_2OA_3=.....\angle A_{n-1}OA_n=\frac{2\pi}{n}$$
Therefore each of these points can be written as, $$A_k=ce^{i(k-1)\frac{2\pi}{n}}\quad\forall\;k=1,2,3...n$$
Now I can write $$r_k=|re^{i\theta}-ce^{i(k-1)\frac{2\pi}n}|$$
Now I was thinking of using $|u||v|=|uv|$, so the product can be written as $$\prod_{k=1\to n} r_k^2=|(re^{i\theta}-c).(re^{i\theta}-ce^{i\frac{2\pi}{n}}).(re^{i\theta}-ce^{i\frac{4\pi}{n}})....(re^{i\theta}-ce^{i(n-1)\frac{2\pi}{n}})|^2$$ $$\implies \prod_{k=1\to n}r_k^2=|r^ne^{n\theta}-r^{n-1}e^{i(n-1)\theta}*c\left(\sum_{k=1\to n}e^{i(k-1)\frac{2\pi}{n}}\right)+r^{n-2}e^{i(n-2)\theta}*c^2\left(\sum_{k_1=1\to n, k_2=k_1+1\to n}e^{i(k_1-1)\frac{2\pi}{n}}e^{i(k_2-1)\frac{2\pi}{n}}\right)+.....(-1)^nc^n\prod_{k=1\to n}e^{i(k-1)\frac{2\pi}{n}}|^2$$
Now, how do I proceed from here? Do I have to reduce all the exponential in brackets to simpler forms. For example some are coming out to be simple,
$$\begin{align} \sum_{k=1\to n}e^{i(k-1)\frac{2\pi}{n}} & =1+e^{i\frac{2\pi}{n}}+e^{i\frac{4\pi}{n}}+e^{i\frac{6\pi}{n}}+.....e^{i\frac{(n-1)2\pi}{n}} \\ & = \frac{1*\big((e^{i\frac{2\pi}{n}})^n-1)}{e^{i\frac{2\pi}{n}}-1} \\ & = 0 \end{align}$$
or the last one
$$y=\prod_{k=1\to n}e^{i(k-1)\frac{2\pi}{n}}$$ $$ \begin{align} ln\,y &= 0+i\frac{2\pi}{n}+i\frac{4\pi}{n}+....i\frac{(n-1)2\pi}{n}\\ &=i\frac{2\pi}{n}(1+2+3+...(n-1))\\ &=i\frac{2\pi}{n}*\frac{(n-1)n}{2}\\ &=i(n-1)\pi \end{align}$$ $$ \implies y=e^{(i(n-1)\pi)}$$
Honestly, this seems to be a very tedious process, which is why I think I am wrong in my approach. Any help is appreciated
