Prove that
\begin{equation} \prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)= \left\{ \begin{aligned} \sqrt{n} \space \space \text{for $n$ odd}\\ \\ \ 1 \space \space \text{for $n$ even}\\ \end{aligned} \right. \end{equation}
I found this identity here at$(35)$. At the moment I don't know where I should start from. Thanks!
$$\tan nx=\frac{^nC_1t-^nC_3t^3+^nC_5t^5-\cdots }{^nC_0t^0-^nC_2t^2+^nC_4t^4-\cdots }$$ where $t=\tan x$
If $\tan nx=0, x=\frac {k\pi}n$ where $0\le k< n$, clearly, the roots of this $n$-degree equation are $\tan\frac{k\pi}n$
If $n$ is odd,
$^nC_n(-1)^{\frac{n-1}2}t^n+^nC_{n-2}(-1)^{\frac{n-3}2}t^{n-2}+\cdots-^nC_3t^3+^nC_1t=0$
$^nC_n(-1)^{\frac{n-1}2}t^{n-1}+^nC_{n-2}(-1)^{\frac{n-3}2}t^{n-3}+\cdots-^nC_3t^2+^nC_1=0$ if we exclude $k=0$
So, $\prod_{k=1}^{n-1}\tan \left(\frac{k \pi}{n}\right)=n(-1)^{\frac{n-1}2}$ (applying Vieta's formula)
Now, $\tan \left(\frac{(n-k) \pi}{n}\right)=\tan \left(\pi-\frac{k \pi}{n}\right)=-\tan \left(\frac{k \pi}{n}\right)$
So, there are $\frac{n-1}2$ such pairs and $\lfloor \frac{n-1}2 \rfloor=\frac{n-1}2$ as $n$ is odd.
$\prod_{k=1}^{n-1}\tan \left(\frac{k \pi}{n}\right)$ $=(-1)^{\frac{n-1}2}\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan^2 \left(\frac{k \pi}{n}\right) $
$\implies (-1)^{\frac{n-1}2}\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan^2 \left(\frac{k \pi}{n}\right)=n(-1)^{\frac{n-1}2} $
$\implies \left(\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)\right)^2=n$
$\implies \prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)=\sqrt n$ as all the angles lies in $(0,\frac \pi 2)$
If $n$ is even,
$ ^nC_1t^{n-1}-^nC_3t^{n-3}+^nC_5t^{n-5}-\cdots+^nC_{n-1}(-1)^{\frac n 2}t=0$ which has roots $\tan\frac{k\pi}n$ where $0\le k<n$ and $k\ne \frac n 2$ as $k=\frac n 2$ corresponds to $\tan \frac \pi 2(=\infty)$ which has occurred as the co-efficient of $t^n$ is $0$.
So, $ ^nC_1t^{n-2}-^nC_3t^{n-4}+^nC_5t^{n-6}-\cdots+^nC_{n-1}(-1)^{\frac n 2}=0$ if we exclude $k=0$ i.e., $(n-2)$ degree equation in $t$.
So, $$\prod_{\substack{k=1 \\ k\neq \frac{n}{2}}}^{n-1}\tan \left(\frac{k \pi}{n}\right)=-(-1)^{\frac n 2}$$
Now, $\tan \left(\frac{(n-k) \pi}{n}\right)=\tan \left(\pi-\frac{k \pi}{n}\right)=-\tan \left(\frac{k \pi}{n}\right)$
So, there are $\frac{n-2}2=(\frac n 2 -1)$ such pairs and $\lfloor \frac{n-1}2 \rfloor=\frac{n-2}2$ as $n$ is even.
$-(-1)^{\frac n 2}$ $={\displaystyle\prod_{\substack{k=1 \\ k \neq \frac{n}{2}}}^{n-1}} \tan \left(\frac{k \pi}{n}\right)$ $=(-1)^{\frac{n-2}2}\left(\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)\right)^2$
$\implies \left(\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)\right)^2=1$
$\implies \prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)=1$ as all the angles lies in $(0,\frac \pi 2)$
This answer only presents the answers already given, but in what I hope is a more accessible form.
Even n
A key identity is $\tan(x)\tan(\pi/2-x)=1$. For $n$ even, this immediately verifies that $$ \prod_{k=1}^{n/2-1}\tan\left(\frac{k\pi}{n}\right)=1\tag{1} $$ Odd n
Since $\cos(nx)(1+i\tan(nx))=e^{inx}=(\cos(x)(1+i\tan(x)))^n$. Considering the ratio of the imaginary part to the real part, for odd $n$, we get $$ \tan(nx)=\frac{\displaystyle\sum_{k=0}^{(n-1)/2}(-1)^{k}\binom{n}{2k+1}\tan^{2k+1}(x)}{\displaystyle\sum_{k=0}^{(n-1)/2}(-1)^{k}\binom{n}{2k}\tan^{2k}(x)}\tag{2} $$ Therefore, $$ \sum_{k=0}^{(n-1)/2}(-1)^{k}\binom{n}{2k+1}x^{2k}=0\tag{3} $$ has roots at $x\in\left\{\tan\left(\dfrac{k\pi}{n}\right):1\le k\le n-1\right\}$. Since $(3)$ has even degree, the product of the roots is the ratio of the constant coefficient to the lead coefficient: $$ \prod_{k=1}^{n-1}\tan\left(\dfrac{k\pi}{n}\right)=(-1)^{(n-1)/2}n\tag{4} $$ Another key identity is $\tan(\pi-x)=-\tan(x)$. Combined with $(4)$, this yields $$ \prod_{k=1}^{(n-1)/2}\tan\left(\dfrac{k\pi}{n}\right)=\sqrt{n}\tag{5} $$
I forgot that I had posted an answer to this question, and answered a duplicate question recently. Since this answer to the odd case is significantly different from the other answers, I have moved it here.
Note that $$ \tan^2(\theta/2)=-\left(\frac{e^{i\theta}-1}{e^{i\theta}+1}\right)^2 $$ Therefore, for odd $n$, $$ \begin{align} \prod_{k=1}^{(n-1)/2}\tan^2(k\pi/n) &=\prod_{k=1}^{(n-1)/2}(-1)\left(\frac{e^{2\pi ik/n}-1}{e^{2\pi ik/n}+1}\right)^2\\ &=\prod_{k=1}^{(n-1)/2}\left(\frac{e^{2\pi ik/n}-1}{e^{2\pi ik/n}+1}\right)\left(\frac{e^{-2\pi ik/n}-1}{e^{-2\pi ik/n}+1}\right)\\ &=\prod_{k=1}^{(n-1)/2}\left(\frac{e^{2\pi ik/n}-1}{e^{2\pi ik/n}+1}\right)\left(\frac{e^{2\pi i(n-k)/n}-1}{e^{2\pi i(n-k)/n}+1}\right)\\ &=\prod_{k=1}^{n-1}\frac{e^{2\pi ik/n}-1}{e^{2\pi ik/n}+1}\\ &=\prod_{k=1}^{n-1}\frac{1-e^{2\pi ik/n}}{1+e^{2\pi ik/n}}\\ &=\lim_{z\to1}\prod_{k=1}^{n-1}\frac{z-e^{2\pi ik/n}}{z+e^{2\pi ik/n}}\\ &=\lim_{z\to1}\frac{z^n-1}{z-1}\frac{z+1}{z^n+1}\\[12pt] &=n \end{align} $$ Since tangent is positive in the first quadrant, $$ \prod_{k=1}^{(n-1)/2}\tan(k\pi/n)=\sqrt{n} $$
Partial solution:
Note that
$$\tan(\frac{\pi}{2}-x) \tan(x)=1$$
This solves the problem for $n$ even: if $n=2m$, then
$$P:=\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)=\prod_{k=1}^{m-1}\tan \left(\frac{k \pi}{2m}\right) =\prod_{k=1}^{m-1}\tan \left(\frac{(m-k) \pi}{2m}\right) $$
hence
$$P^2=\prod_{k=1}^{m-1}\tan \left(\frac{k \pi}{2m}\right)\prod_{k=1}^{m-1}\tan \left(\frac{(m-k) \pi}{2m}\right)=1$$
and since $P$ is positive, you are done...
