Mysterious connection (2)

More generally, let $n \ge 3$ be a positive integer. Let $\omega = e^{2 \pi i/n}$ , so $\omega$ is a primitive $n$ th root of unity. Then the roots of $x^n - 1 = 0$ are $x = 1$ , $\omega$ , $\omega^2$ , $\dots$ , $\omega^{n - 1}$ , so $x^n - 1 = (x - 1)(x - \omega)(x - \omega^2) \dotsm (x - \omega^{n - 1}).$ Also, $x^n - 1 = (x - 1)(x^{n - 1} + x^{n - 2} + \dots + x + 1),$ so $(x - \omega)(x - \omega^2) \dotsm (x - \omega^{n - 1}) = x^{n - 1} + x^{n - 2} + \dots + x + 1.$ Setting $x = 1$ , we get $(1 - \omega)(1 - \omega^2) \dotsm (1 - \omega^{n - 1}) = n.$

For $1 \le k \le n$ , let $P_k$ be the point corresponding to the complex number $\omega^{k - 1}$ . Then $P_1 P_k = |1 - \omega^{k - 1}|,$ so $\prod_{k = 2}^n P_1 P_k = \prod_{k = 2}^n |1 - \omega^{k - 1}| = \left| \prod_{k = 2}^n (1 - \omega^{k - 1}) \right| = n.$

Note that $\begin{aligned} \sin(2n+1)x & = \; \mathrm{Im}\big((\cos x + i \sin x)^{2n+1}\big) \; = \; \sum_{j=0}^n \binom{2n+1}{2j+1}(-1)^j \cos^{2n-2j}x\sin^{2j+1}x \\ & = \; \sin x \sum_{j=0}^n (-1)^j \binom{2n+1}{2j+1} \sin^{2j}x (1 - \sin^2x)^{n-j} \; = \; \sin x F_n(\sin^2x) \end{aligned}$ where $F_n(X) \; = \; \sum_{j=0}^n(-1)^j\binom{2n+1}{2j+1}X^j(1-X)^{n-j}$ is a polynomial of degree $n$ with leading coefficient $(-1)^n\sum_{j=0}^n \binom{2n+1}{2j+1} =(-1)^n2^{2n}$ and constant term $\binom{2n+1}{1} = 2n+1$ . It is clear from the definition of $F_n$ that the roots of $F_n$ are $\sin^2\tfrac{j\pi}{2n+1}$ for $1 \le j \le n$ , so we deduce that $\prod_{j=1}^n \sin^2\tfrac{j\pi}{2n+1} \; = \; 2^{-2n}(2n+1)$ Thus we deduce that $\prod_{n=2}^{17}P_1P_n \; = \; 2^{16}\prod_{n=1}^{16} \sin \tfrac{n\pi}{17} \; = \; 2^{16}\prod_{n=1}^8 \sin^2\tfrac{n\pi}{17} \; = \; \boxed{17}$