Let $$C = \cos\frac{\pi}{2n+1}\cdot\cos\frac{2\pi}{2n+1}\cdots\cos\frac{n\pi}{2n+1}$$ and $$S = \sin\frac{\pi}{2n+1}\cdot\sin\frac{2\pi}{2n+1}\cdots\sin\frac{n\pi}{2n+1}$$ Then, \begin{align} C\cdot S &=\Big(\cos\frac{\pi}{2n+1}\cdot\cos\frac{2\pi}{2n+1}\cdots\cos\frac{n\pi}{2n+1}\Big)\Big(\sin\frac{\pi}{2n+1}\cdot\sin\frac{2\pi}{2n+1}\cdots\sin\frac{n\pi}{2n+1}\Big)\\ &=\Big(\sin\frac{\pi}{2n+1}\cos\frac{\pi}{2n+1}\Big)\Big(\sin\frac{2\pi}{2n+1}\cos\frac{2\pi}{2n+1}\Big)\cdots\Big(\sin\frac{n\pi}{2n+1}\cos\frac{n\pi}{2n+1}\Big)\\ &=\Big(\frac{1}{2}\cdot\sin\frac{2\pi}{2n+1}\Big)\Big(\frac{1}{2}\cdot\sin\frac{4\pi}{2n+1}\Big)\cdots\Big(\frac{1}{2}\cdot\sin\frac{2n\pi}{2n+1}\Big)\\ &=\frac{1}{2^n}\cdot S \end{align} $$\therefore\quad C=\boxed{\frac{1}{2^n}}$$