For natural $n$, prove $\prod_{k=1}^n \tan\left(\frac{k\pi}{2n+1}\right) = 2^n \prod_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}$ 
Prove that, for a natural number $n$,
  $$\prod_{k=1}^n \tan\left(\frac{k\pi}{2n+1}\right) = 2^n \prod_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}$$

This follows from a continued fraction identity for which, I think, there is a lengthy proof. But, I thought, that there may be a direct geometric or another proof. Constructing a polynomial with the sines and tangents roots may be helpful.
 A: The proof for this should be identical to the one for:
$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$$
A: The products for $\sin(x)$ and $\cos(x)$ can be evaluated in the same manner. Taking the ratio gives the product for $\tan(x)$.
$$
\begin{align}
\prod_{k=1}^n\sin\left(\frac{k\pi}{2n+1}\right)
&=\prod_{k=1}^n\left|\frac{e^{i\frac{k\pi}{2n+1}}-e^{-i\frac{k\pi}{2n+1}}}{2i}\right|\tag{1a}\\
&=\frac1{2^n}\prod_{k=1}^n\left|1-e^{-i\frac{2k\pi}{2n+1}}\right|\tag{1b}\\
&=\frac1{2^n}\prod_{\substack{k=-n\\k\ne0}}^n\left|1-e^{-i\frac{2k\pi}{2n+1}}\right|^{1/2}\tag{1c}\\
&=\lim_{z\to1}\frac1{2^n}\prod_{\substack{k=-n\\k\ne0}}^n\left|z-e^{-i\frac{2k\pi}{2n+1}}\right|^{1/2}\tag{1d}\\
&=\lim_{z\to1}\frac1{2^n}\left|\frac{z^{2n+1}-1}{z-1}\right|^{1/2}\tag{1e}\\[6pt]
&=\frac{\sqrt{2n+1\vphantom{-}}}{2^n}\tag{1f}
\end{align}
$$
Explanation:
$\text{(1a):}$ $\sin(x)\gt0$ over the product
$\phantom{\text{(1a):}}$ write $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$\text{(1b):}$ $|i|=\left|e^{i\frac{k\pi}{2n+1}}\right|=1$
$\text{(1c):}$ $|1-z|=|1-\bar z|$
$\text{(1d):}$ write $1=\lim\limits_{z\to1}z$
$\text{(1e):}$ $\prod\limits_{\substack{k=-n\\k\ne0}}^n\left(z-e^{-i\frac{2k\pi}{2n+1}}\right)=\frac{z^{2n+1}-1}{z-1}$
$\text{(1f):}$ evaluate the limit
$$
\begin{align}
\prod_{k=1}^n\cos\left(\frac{k\pi}{2n+1}\right)
&=\prod_{k=1}^n\left|\frac{e^{i\frac{k\pi}{2n+1}}+e^{-i\frac{k\pi}{2n+1}}}2\right|\tag{2a}\\
&=\frac1{2^n}\prod_{k=1}^n\left|1+e^{-i\frac{2k\pi}{2n+1}}\right|\tag{2b}\\
&=\frac1{2^n}\prod_{\substack{k=-n\\k\ne0}}^n\left|1+e^{-i\frac{2k\pi}{2n+1}}\right|^{1/2}\tag{2c}\\
&=\lim_{z\to1}\frac1{2^n}\prod_{\substack{k=-n\\k\ne0}}^n\left|z+e^{-i\frac{2k\pi}{2n+1}}\right|^{1/2}\tag{2d}\\
&=\lim_{z\to1}\frac1{2^n}\left|\frac{z^{2n+1}+1}{z+1}\right|^{1/2}\tag{2e}\\[6pt]
&=\frac1{2^n}\tag{2f}
\end{align}
$$
Explanation:
$\text{(2a):}$ $\cos(x)\gt0$ over the product
$\phantom{\text{(2a):}}$ write $\cos(x)=\frac{e^{ix}+e^{-ix}}2$
$\text{(2b):}$ $\left|e^{i\frac{k\pi}{2n+1}}\right|=1$
$\text{(2c):}$ $|1+z|=|1+\bar z|$
$\text{(2d):}$ write $1=\lim\limits_{z\to1}z$
$\text{(2e):}$ $\prod\limits_{\substack{k=-n\\k\ne0}}^n\left(z+e^{-i\frac{2k\pi}{2n+1}}\right)=\frac{z^{2n+1}+1}{z+1}$
$\text{(2f):}$ evaluate the limit
Thus, taking the ratio of $(1)$ and $(2)$ gives
$$
\prod_{k=1}^n\tan\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1\vphantom{-}}\tag3
$$
