I am looking to evaluate
$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n}$$
without using complex numbers. I can show the result if $n$ is a power of $2$, but if $n$ is anything else I reach a point where I can no longer proceed.
My current method (for when $n$ is a power of $2$) is to note that all of the angle that sin is being evaluated on are evenly distributed between $0$ and $\pi$, but more importantly, one of the angles will always be $\frac{\pi}{2}$ and all angles which are greater than $\frac{\pi}{2}$ can be re-written as the cos of an angle that is less than $\frac{\pi}{2}$ (using the complementary angle). When this is done, we get
$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \prod_{k = 1}^{\lfloor\frac{n - 1}{2}\rfloor}\frac{1}{2}\sin\frac{2k\pi}{n} = \frac{1}{2^{\lfloor\frac{n - 1}{2}\rfloor}}\prod_{k = 1}^{\lfloor\frac{n - 1}{2}\rfloor}\sin\frac{k\pi}{\frac{n}{2}}$$
This process can then be repeated recursively until we have just one $\sin$ term left in the product (which will be $\sin\frac{\pi}{2}$), then what is left is equivalent to
$$\frac{n}{2^{n - 1}}\;.$$
Can anyone suggest a way of generalizing this to when $n$ is not a power of $2$?
Thanks in advance.