A deduction from $x^n-1=(x^2-1)\prod_{k=1}^{(n-2)/2}[x^2-2x\cos{\frac{2k\pi}{n}}+1]$ Prove that $$x^n-1=(x^2-1)\prod_{k=1}^{(n-2)/2}[x^2-2x\cos{\frac{2k\pi}{n}}+1]$$ if $n$ be an even positive integer. Hence deduce that $$\sin{\frac{\pi}{32}}\sin{\frac{2\pi}{32}}\sin{\frac{3\pi}{32}}\cdots\sin{\frac{15\pi}{32}}=\frac{1}{2^{13}}$$
Attempt:
I am able to solve the 1st part of the problem and unable to solve the 2nd part that is the deduction part of the problem. Please help for the deduction part only. 
Edit:
Using double angle: I can show that $$\sqrt{n}=2^{(n-1)/2}\sin{\frac{\pi}{n}}\sin{\frac{2\pi}{n}}\cdots \sin{\frac{(n-2)\pi}{2n}}$$
Please suggest suitable n and the please help me to complete the remaining part. I have taken n=32 but not able to complete the solution.

Hint $x=\frac{\pi}{32}$ so we can write it as $(\ sinx\sin(15x))(\sin(2x)\sin(14x))...sin(8x)$ now multiply/divide by $2^7$ and we know $sin(8\pi/32)=1/\sqrt{2}$  then after multiplying we get $\frac{1}{2^7\sqrt{2}}.(\cos(\pi/2)-\cos(7\pi/16)).....(\cos(0)-\cos(\pi/16))$ as $2\sin(a)\sin(b)=\cos(b-a)-\cos(a+b)$ so we get $\frac{1}{2^8}cos(7x)cos(6x)cos(5x)cos(3x)cos(2x)cos(x)$ now again multiply/divide $2^3$ and use $2\cos(a)\cos(b)=\cos(a+b)+cos(a-b)$ then we get $\frac{1}{2^{11}}\cos(3\pi/8)\cos(\pi/4)cos(\pi/8)$ and getting result from here is easy or you can use sin as $R(e^{ix})$ So we can write it as $$\sum e^{inx}$$ where $n\in [1,8]$ and then you have geometric series which can be calculated but maybe it will require a calculator to verify final result
