Stuck trying to sum this series I actually got this question from a book - TMH. Here is the question:

Prove that $\cot^2 \frac{\pi}{2n+1}\cdot \cot^2 \frac{\pi}{2n+1}\cdot \ldots \cdot \cot^2 \frac{\pi}{2n+1}$ are the roots of the equation
$$
x^n - \frac{\binom{2n+1}{3}}{\binom{2n+1}{1}} x^{n-1} +\frac{\binom{2n+1}{5}}{\binom{2n+1}{1}}x^{n-2}-\cdots = 0,
$$and hence show that $\lim\limits_{n\to\infty} \sum_{r=1}^{n} r^{-2}=\frac{\pi^2}{6}$.

I noticed the following, but couldn't solve later.
$$
\cot ^{2} \frac{\pi}{2 n+1}=\tan ^{2} \frac{n \pi}{2 n+1}
$$

You may know the following formula that you will find for example here in paragraph "Sine, cosine, tangent of multiple angles"
$$\displaystyle \tan(n\theta )=\frac {\sum _{k{\text{ odd}=2q+1}}(-1)^{q}{n \choose k}(\tan\theta) ^{2q+1}}{\sum _{k{\text{ even}=2q}}(-1)^{q}{n \choose k}(\tan\theta)^{2q} }\tag{1}$$
Let us take $2n+1$ instead of $n$ in (1) :
$$\displaystyle \tan((2n+1)\theta )=\frac {\sum _{q=0}^n(-1)^{q}{2n+1 \choose 2q+1}(\tan\theta) ^{2q+1}}{\sum _{q=0}^{n}(-1)^{q}{2n+1 \choose 2q}(\tan\theta)^{2q} }\tag{2}$$
Therefore, we have 
$$\tan((2n+1)\theta)=0 \ \ \iff \ \ \sum _{q=0}^n (-1)^{q}{2n+1 \choose 2q+1}x^q = 0\ \ \text{with} \ \ x:=\tan(\theta)^2\tag{3}$$
Two remarks :
1) The second equation in (3) is exactly your equation (multiplied by the common denominator) ?
2) This equation has been simplified by $x$ assumed $\ne 0$.
The first equation in (3) is equivalent to : 
$$(2n+1) \theta=k\dfrac{\pi}{2} \ \ \text{for any integer} \ k \tag{4}$$
Therefore, $\theta$ can take the values : 
$$\theta=\dfrac{k}{2n+1}\dfrac{\pi}{2}\tag{5}$$
Plugging (5) into relationship $x=(\tan\theta)^2$, we get
$$x=\tan(\dfrac{k}{2n+1}\dfrac{\pi}{2})^2=(\cot(\dfrac{\pi}{2}-\dfrac{k}{2n+1}\dfrac{\pi}{2}))^2=(\cot(\dfrac{2n+1-k}{2n+1}\dfrac{\pi}{2}))^2$$
as desired.
