Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$ 
Question : Is the following true for any $m\in\mathbb N$?
  $$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$

Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it. Can anyone help?
By the way, I've been able to prove $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{{\pi}^2}{6}$ by using $(\star)$.
Proof : Let 
$$f(x)=\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{(x-\sin x)(x+\sin x)}{x^2\sin^2 x}.$$
We know that $f(x)\gt0$ if $0\lt x\le {\pi}/{2}$, and that $\lim_{x\to 0}f(x)=1/3$. Hence, letting $f(0)=1/3$, we know that $f(x)$ is continuous and positive at $x=0$. Hence, since $f(x)\ (0\le x\le {\pi}/2)$ is bounded, there exists a constant $C$ such that $0\lt f(x)\lt C$. Hence, substituting $x={(k\pi)}/{(2n+1)}$ for this, we get
$$0\lt \frac{1}{\frac{2n+1}{{\pi}^2}\sin^2\frac{k\pi}{2n+1}}-\frac{1}{k^2}\lt\frac{{\pi}^2C}{(2n+1)^2}.$$
Then, the sum of these from $1$ to $n$ satisfies 
$$0\lt\frac{{\pi}^2\cdot 2n(n+1)}{(2n+1)^2\cdot 3}-\sum_{k=1}^{n}\frac{1}{k^2}\lt\frac{{\pi}^2Cn}{(2n+1)^2}.$$
Here, we used $(\star)$. Then, considering $n\to\infty$ leads what we desired.

There not being a residue theory answer, I have copied excerpts from this answer:
$\frac{2n/z}{z^{2n}-1}$ has residue $1$ at $z=e^{\pi ik/n}$ and residue $-2n$ at $z=0$.
Thus,
$$
\begin{align}
\left(\frac{2i}{z-\frac1z}\right)^2\frac{2n/z}{z^{2n}-1}
&=\frac{-4z^2}{z^4-2z^2+1}\frac{2n/z}{z^{2n}-1}\\
&=\left(\frac1{(z+1)^2}-\frac1{(z-1)^2}\right)\frac{2n}{z^{2n}-1}
\end{align}
$$
has residue $\csc^2\left(\pi k/n\right)$ at $z=e^{\pi ik/n}$ except at $z=\pm1$. A bit of computation gives
$$
\begin{align}
\frac{2n}{z^{2n}-1}
&=\phantom{+}\frac1{z-1}-\frac{2n-1}2+\frac{(2n-1)(2n+1)}{12}(z-1)+O\!\left((z-1)^2\right)\\
&=-\frac1{z+1}-\frac{2n-1}2-\frac{(2n-1)(2n+1)}{12}(z+1)+O\!\left((z+1)^2\right)
\end{align}
$$
Therefore,
$$\newcommand{\Res}{\operatorname*{Res}}
\Res_{z=1}\left(\frac1{(z+1)^2}-\frac1{(z-1)^2}\right)\frac{2n}{z^{2n}-1}
=\frac14-\frac{4n^2-1}{12}=-\frac{n^2-1}3
$$
and
$$
\Res_{z=-1}\left(\frac1{(z+1)^2}-\frac1{(z-1)^2}\right)\frac{2n}{z^{2n}-1}
=\frac14-\frac{4n^2-1}{12}=-\frac{n^2-1}3
$$
Since the sum of the residues at all the singularities is $0$, we get that half the sum over the singularities except at $z=\pm1$ is
$$
\sum_{k=1}^{n-1}\csc^2\left(\frac{k}{n}\pi\right)=\frac{n^2-1}3
$$
