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 

$\text{We are going to prove that}$
$\displaystyle \prod_{k=1}^{m}\left(\tan \frac{k \pi}{2 m+1}\right)=\sqrt{2 m+1}. \tag*{}$
$\textrm{Considering the roots of the polynomial equation }\\ \text{of degree 2m}$
$\displaystyle (z+1)^{2 m+1}-(z-1)^{2 m+1}=0 \cdots(*)\tag*{} $
$\displaystyle \left(\frac{z-1}{z+1}\right)^{2 m+1}=1 \tag*{}  \\$
$\displaystyle \frac{z-1}{z+1} =\cos \frac{2 k \pi}{2 m+1}+i \sin \frac{2 k \pi}{2 m+1} =e^{\frac{2 k \pi}{2 m+1} i}, $
$\text { where } k=1,2, \cdots , 2 m.$
$\textrm{Hence the 2m complex roots of (*) are}$
$ \displaystyle z =\frac{e^{\frac{2 k \pi}{2 m+1} i}+1}{1-e^{\frac{2 k \pi}{2 m+1} i}} =\frac{e^{\frac{k \pi}{2 m+1} i}-e^{-\frac{k \pi}{2 m+1} i}}{e^{-\frac{k\pi}{2 m+1} i}-e^{\frac{k \pi}{2 m+1} i}} \tag*{} $
$\displaystyle \qquad  =\frac{2 \cos \left(\frac{k \pi}{2 m+1}\right)}{-2 i\sin \left(\frac{k \pi}{2 m+1}\right)}  =i \cot \left(\frac{k \pi}{2 m+1}\right)\tag*{}  $
$\displaystyle \because \text { product of roots of }(*) =\frac{\text{constant in (*)}}{\text{leading coefficient of (*)}}\tag*{}  $
$\therefore \displaystyle \prod_{k=1} ^{2 m}\left[i \cot \left(\frac{k \pi}{2 m+1}\right)\right]=\frac{2}{2(2 m+1)} \tag*{} \\$
$\displaystyle i^{2 m} \prod_{k=1}^ {2 m}\cot \left(\frac{k \pi}{2 m+1}\right)=\frac{1}{ 2 m+1} \tag*{} $
$\because \displaystyle \cot \frac{(2 m+1-k)\pi}{2 m+1} =-\cot \left(\frac{k \pi}{2 m+1}\right)\tag*{} $
$\therefore\displaystyle (-1)^{m}(-1) ^m  \prod_{k=1}^ { m}\left[ \cot ^{2} \frac{k \pi}{2 m+1}\right]=\frac{1}{2 m+1} \tag*{} $
$\displaystyle \left(\prod_{k=1}^{m} \cot \frac{k \pi}{2 m+1}\right)^{2}=\frac{1}{2 m +1 } \tag*{}  \\$
$$\displaystyle \quad \prod_{k=1}^{m}\left(\tan \frac{k \pi}{2 m+1}\right)=\sqrt{2 m+1}$$
