Proving that $\prod_{k=1}^{n}\tan\left(\frac{\pi k}{2n+1}\right)=\sqrt{2n+1}$ using geometry I have to prove that the following holds. A hint to use complex numbers has been given. I have tried to make a start but not to any result. 

$$\prod_{k=1}^{n}\tan\left(\frac{\pi k}{2n+1}\right)=\sqrt{2n+1}$$


My Attempt:
Let us consider $(2n+1)^{\text{th}}$ roots of unity, $z_k=\exp\left(\frac{2k\pi i}{2n+1}\right)$. We can rewrite the product in terms of $\arg(z_k)$ as $\prod_{k=1}^{n}\tan\left(\frac{1}{2}\arg(z_k)\right)$. Or equivalently so, if we consider $(4n+2)^{\text{th}}$ roots of unity, we get this product as $\prod_{k=1}^{n}\tan(\arg(\zeta _k))$, where $\zeta_k= \exp\left(\frac{2k\pi i}{4n+2}\right)$. 

I know it can be proved by proving the expression for $\prod\sin(\frac{1}{2}\arg(z_k))$. But I was wondering, is there a way to use telescopic products or purely the geometry of the complex roots of unity to arrive at this 

Building on @WETutorialSchool's hint, start from the identity$$i\tan\frac{\theta}{2}=\frac{2i\sin\frac{\theta}{2}\exp\frac{i\theta}{2}}{2\cos\frac{\theta}{2}\exp\frac{i\theta}{2}}=\frac{\exp i\theta-1}{\exp i\theta+1}.$$Define $z:=\exp\frac{2i\pi}{2n+1}$ so$$\frac{z^k-1}{z^k+1}=i\tan\frac{k\pi}{2n+1},\,\frac{z^{2n+1-k}-1}{z^{2n+1-k}+1}=\frac{z^{-k}-1}{z^{-k}+1}=\frac{1-z^k}{1+z^k}=-i\tan\frac{k\pi}{2n+1}.$$Since $\prod_{k=1}^n\tan\frac{k\pi}{2n+1}$ is a product of the positive tangents of $n$ acute angles,$$\prod_{k=1}^n\tan\frac{k\pi}{2n+1}=\sqrt{\prod_{k=1}^ni\tan\frac{k\pi}{2n+1}\cdot-i\tan\frac{k\pi}{2n+1}}=\sqrt{\prod_{k=1}^n\frac{z^k-1}{z^k+1}\frac{z^{2n+1-k}-1}{z^{2n+1-k}+1}}=\sqrt{\prod_{k=1}^{2n}\frac{z^k-1}{z^k+1}}.$$To prove $\prod_{k=1}^{2n}\frac{z^k-1}{z^k+1}=2n+1$, consider the values of $z^k+1,\,0\le k\le 2n$. They are the roots of $(w-1)^{2n+1}-1$, so their product is $-1$ times this polynomial's constant term, i.e. $2$. In other words, $\prod_{k=0}^{2n}(z^k+1)=2$ and $\prod_{k=1}^{2n}(z^k+1)=1$. Similarly, $\prod_{k=1}^{2n}(z^k-1)$ is the product of the roots of $\frac{(w+1)^{2n+1}-1}{w}$, so is equal to its constant term, $\binom{2n+1}{1}$.
