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.

Fixing $n=32$, we can rewrite the equation like this:
$$\frac{x^{32}-1}{x^2-1} = \prod_{k=1}^{15}\left(x^2+1-2x\cos\frac{k\pi}{16}\right)$$
for all $x\neq\pm1$.
We will take the limit as $x\to 1$ of both sides of the equality.
First, we have (using L'Hôpital's rule)
$$\lim\limits_{x\to 1} \frac{x^{32}-1}{x^2-1} = \lim\limits_{x\to 1}\frac{32x^{31}}{2x} = 16$$
Then, plugging in $x=1$, we have
$$x^2+1-2x\cos\frac{k\pi}{16} = 2-2\cos\frac{k\pi}{16} = 4\cdot\frac{1-\cos(k\pi/16)}{2} = 4\sin^2\frac{k\pi}{32}$$
The product becomes
$$\prod_{k=1}^{15}4\sin^2\frac{k\pi}{32}= 4^{15}\left(\prod_{k=1}^{15}\sin\frac{k\pi}{32}\right)^2$$
Note that the limit as $x\to 1$ of the product is equal to its value at $x=1$ because the expression is continuous (it is the product of continuous functions).
Now, equating the limits we get
$$16 = 4^{15}\left(\prod_{k=1}^{15}\sin\frac{k\pi}{32}\right)^2$$
As all the factors in the product are positive, we can take the square root of both sides and we get
$$\sin \frac{\pi}{32}\sin \frac{2\pi}{32}\sin \frac{3\pi}{32}\cdots\sin \frac{15\pi}{32}= \prod_{k=1}^{15}\sin\frac{k\pi}{32} =\frac{1}{2^{13}}$$
