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.

The roots of the polynomial $X^{2n}-1$ are $\omega_j:=\exp\left(\mathrm i\frac{2j\pi}{2n}\right)$, $0\leq j\leq 2n-1$. We can write 
\begin{align}
X^{2n}-1&=(X^2-1)\prod_{j=1}^{n-1}\left(X-\exp\left(\mathrm i\frac{2j\pi}{2n}\right)\right)\left(X-\exp\left(-\mathrm i\frac{2j\pi}{2n}\right)\right)\\
&=(X^2-1)\prod_{j=1}^{n-1}\left(X^2-2\cos\left(\frac{j\pi}n\right)X+1\right).
\end{align}
Evaluating this at $X=i$, we get 
$$(-1)^n-1=(-2)(-2\mathrm i)^{n-1}\prod_{j=1}^{n-1}\cos\left(\frac{j\pi}n\right),$$
hence 
\begin{align}
\prod_{j=1}^n\cos\left(\frac{j\pi}n\right)&=-\prod_{j=1}^{n-1}\cos\left(\frac{j\pi}n\right)\\
&=\frac{(-1)^n-1}{2(-2\mathrm i)^{n-1}}\\
&=\frac{(-1)^n-1}2\cdot \frac{\mathrm i^{n-1}}{2^{n-1}}.
\end{align}
The RHS is $0$ if $n$ is even, and $-\dfrac{(-1)^m}{2^{2m}}=-\dfrac{\sin(n\pi/2)}{2^{n-1}}$ if $n$ is odd with $n=2m+1$.
