In my textbook there is a proof for the following
If $n=2m+1$ with $m\in\mathbb N$, then we can write $$F_n(z)=\frac1{2i}\left(\left(1+\frac{iz}n\right)^n-\left(1-\frac{iz}n\right)^n\right)=z\prod_{k=1}^m\left(1-\frac{z^2}{n^2\tan^2(k\pi/n)}\right)$$
In proof we have found $2m+1=n$ distinct roots $z_k$ of $F_n(z)$ where $$z_k=n\tan\left(\frac{k\pi}n\right)$$ and $z_{-m}<z_{-m+1}<...<z_{-1}<z_0<z_1<...<z_{m-1}<z_m$, $z_{-k}=-z_k,z_0=0$ so $$F_n(z)=z\prod_{k=1}^m(z^2-z_k^2)=z\prod_{k=1}^m\left(z^2-n^2\tan^2(\frac{k\pi}n)\right)$$ Now author says "Factoring out $-n^2\tan^2(k\pi/n)$ terms and gathering them all into one constant we can conclude that $$F_n(z)=az\prod_{k=1}^m\left(1-\frac{z^2}{n^2\tan^2(k\pi/n)}\right)$$ for some constant $a$." Then author says $F_n(z)=\frac1{2i}\left(\left(1+\frac{iz}n\right)^n-\left(1-\frac{iz}n\right)^n\right)$ implies that $a=1$. How? Is $a=\prod_{k=1}^m(-n^2\tan^2(k\pi/n))$?
We have
$$\begin{align} F_n(z) &= \frac{1}{2i}\left(\left(1 + \frac{iz}{n}\right)^n - \left(1-\frac{iz}{n}\right)^n\right)\\ &= \frac{1}{2i}\left(\left(1 + iz + O(z^2)\right) - \left(1 - iz+ O(z^2)\right)\right)\\ &= z + O(z^2). \end{align}$$
The coefficient of $z$ on the left is $1$, and on the right in the last formula, the coefficient is $a$.
However, the formula
$$F_n(z) = z\prod_{k=1}^m (z^2-z_k^2)$$
is not correct, there's a constant factor missing. The coefficient of $z^n$ on the right is $1$, but on the left it is
$$\frac{1}{2i}\left(\left(\frac{i}{n}\right)^n - \left(\frac{-i}{n}\right)^n\right) = \frac{1}{2in^n}\left(i^{2m+1} - (-i)^{2m+1}\right) = \frac{2i^{2m+1}}{2in^n} = \frac{(-1)^m}{n^n}.$$
So we have
$$\frac{1}{2i}\left(\left(1 + \frac{iz}{n}\right)^n - \left(1 - \frac{iz}{n}\right)^n\right) = \frac{(-1)^m}{n^n}z\prod_{k=1}^m \left(z^2-n^2\tan^2\left(\frac{k\pi}{n}\right)\right),$$
since both sides are polynomials of degree $n$ with the same zeros and the same leading coefficient. Now factoring out the $-n^2\tan^2\left(\frac{k\pi}{n}\right)$ leads to
$$\begin{align} \frac{1}{2i}\left(\left(1 + \frac{iz}{n}\right)^n - \left(1 - \frac{iz}{n}\right)^n\right) &= \frac{(-1)^m}{n^n}\prod_{k=1}^m \left(-n^2\tan^2\left(\frac{k\pi}{n}\right)\right)z \prod_{k=1}^m \left(1 - \frac{z^2}{n^2\tan^2\frac{k\pi}{n}}\right)\\ &= \frac{1}{n}\prod_{k=1}^m \tan^2\left(\frac{k\pi}{n}\right)\cdot z\prod_{k=1}^m \left(1 - \frac{z^2}{n^2\tan^2\frac{k\pi}{n}}\right), \end{align}$$
and comparing the coefficient of $z$, which is $1$ on the left and
$$\frac{1}{n}\prod_{k=1}^m \tan^2\left(\frac{k\pi}{n}\right)$$
on the right proves
$$\prod_{k=1}^m \tan \left(\frac{k\pi}{2m+1}\right) = \sqrt{2m+1}.$$