How to deduce the following complex number problem I am stuck with the following problem that says:

Using the result $$x^n-1=(x^2-1)\prod_{k=1}^{(n-2)/2}[x^2-2x\cos \frac{2k\pi}{n}+1],$$ if $n$ be an even positive integer,  deduce that $$\sin \frac{\pi}{32}\sin \frac{2\pi}{32}\sin \frac{3\pi}{32}.........\sin \frac{15\pi}{32}=\frac{1}{2^{13}}$$

Can someone point me in the right direction with some explanation? Thanks in advance for your time.

This doesn't use the given information, but it does give some intuition of what you are trying to show.
Consider $x^8 - 1$
The roots of unity are equally spaced around the unit circle. 

$x^8-1 = (x-1)(x-r_1)(x-r_2)\cdots(x-r_7) = (x-1)(x^7 + x^6 + \cdots + 1)$
with $r_k = (\cos \frac {2k\pi}{8}, i\sin \frac{2k\pi}{8})$
$|(x^7 + x^6 + \cdots + 1)| = |x-r_1||x-r_2|\cdots|x-r_7|$
letting $x = 1$
$8 = |1-r_1||1-r_2|\cdots|1-r_7|$
$|1 - r_k| = 2\sin\frac {k\pi}{8}$  (highlighted in the figure for the case $k=3$)
$8 = 2^2\sin\frac {\pi}{8}\sin\frac {2\pi}{8}\sin\frac {5\pi}{8}\cdots\sin\frac {7\pi}{8}\\
\sin\frac {\pi}{8}\sin\frac {2\pi}{8}\cdots\sin\frac {7\pi}{8} = 2^{-4}$
Not a whole lot changes when we consider $x^{32} - 1$
