Evaluating the product $\prod\limits_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)$ Recently, I ran across a product that seems interesting.
Does anyone know how to get to the closed form:
$$\prod_{k=1}^{n}\cos\left(\frac{k\pi}{n}\right)=-\frac{\sin(\frac{n\pi}{2})}{2^{n-1}}$$
I tried using the identity $\cos(x)=\frac{\sin(2x)}{2\sin(x)}$ in order to make it "telescope" in some fashion, but to no avail. But, then again, I may very well have overlooked something.
This gives the correct solution if $n$ is odd, but of course evaluates to $0$ if $n$ is even. 
So, I tried taking that into account, but must have approached it wrong. 
How can this be shown? Thanks everyone.

If $n$ is even, then the term with $k=n/2$ makes the product on the left $0$ and $\sin\left(\frac{n}{2}\pi\right)=0$. So assume that $n$ is odd.
$$
\begin{align}
\prod_{k=1}^n\cos\left(\frac{k\pi}{n}\right)
&=-\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{n}\right)\tag{1}\\
&=-\prod_{k=1}^{n-1}\frac{\sin\left(\frac{2k\pi}{n}\right)}{2\sin\left(\frac{k\pi}{n}\right)}\tag{2}\\
&=\frac{-1}{2^{n-1}}\frac{\prod\limits_{k=\frac{n+1}{2}}^{n-1}\sin\left(\frac{2k\pi}{n}\right)}{\prod\limits_{k=1}^{\frac{n-1}{2}}\sin\left(\frac{(2k-1)\pi}{n}\right)}\tag{3}\\
&=\frac{(-1)^{\frac{n+1}{2}}}{2^{n-1}}\frac{\prod\limits_{k=1}^{\frac{n-1}{2}}\sin\left(\frac{(2k-1)\pi}{n}\right)}{\prod\limits_{k=1}^{\frac{n-1}{2}}\sin\left(\frac{(2k-1)\pi}{n}\right)}\tag{4}\\
&=-\frac{\sin\left(n\frac\pi2\right)}{2^{n-1}}\tag{5}
\end{align}
$$
$(1)$: $\cos(\pi)=-1$
$(2)$: $\sin(2x)=2\sin(x)\cos(x)$
$(3)$: cancel $\sin\left(\frac{j\pi}{n}\right)$ in the numerator and denominator for even $j$ from $2$ to $n-1$
$(4)$: in the numerator, change variable $k\mapsto k+\frac{n-1}{2}$ and use $\sin(x+\pi)=-\sin(x)$
$(5)$: for odd $n$, $\sin\left(n\frac\pi2\right)=(-1)^{\frac{n-1}{2}}$
