sequences and series - Proof of $sum_{n=1}^{infty}frac{1}{n^2}=frac{pi^2}{6}$

Prove that $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$

I have actually got the HINT for this Proof from a very nice book:

The HINT i started with is:

$$\frac{1}{\sin^2 x}=\frac{1}{4\sin^2\left(\frac{x}{2}\right)\cos^2\left(\frac{x}{2}\right)}=\frac{1}{4}\left(\frac{1}{\sin^2\left(\frac{x}{2}\right)}+\frac{1}{\sin^2\left(\frac{x+\pi}{2}\right)}\right) \tag{1}$$

Now we have:

$$1=\frac{1}{\sin^2\left(\frac{\pi}{2}\right)}=\frac{1}{4}\left(\frac{1}{\sin^2\left(\frac{\pi}{4}\right)}+\frac{1}{\sin^2\left(\frac{3\pi}{4}\right)}\right)$$

Applying $(1)$ Repeatedly we get

$$1=\frac{1}{16}\left (\frac{1}{\sin^2\left(\frac{\pi}{8}\right)}+\frac{1}{\sin^2\left(\frac{3\pi}{8}\right)}+\frac{1}{\sin^2\left(\frac{5\pi}{8}\right)}+\frac{1}{\sin^2\left(\frac{7\pi}{8}\right)}\right)$$

So Generalizing we get:

$$1=\frac{1}{4^n}\sum_{k=0}^{2^n-1} \frac{1}{\sin^2\left(\frac{(2k+1)\pi}{2^{n+1}}\right)}$$

EDIT:

Ok let me explain the next step the author of the book has followed:

$$1=\frac{1}{4^n}\sum_{k=0}^{2^n-1} \frac{1}{\sin^2\left(\frac{(2k+1)\pi}{2^{n+1}}\right)}=\frac{2}{4^n}\sum_{k=0}^{2^{n-1}-1} \frac{1}{\sin^2\left(\frac{(2k+1)\pi}{2^{n+1}}\right)}$$

My doubt is how a factor of $2$ came outside and why the upper limit has changed?

Answer

This answer is limited to the question asked in the OP's edit. The puzzling step taken there uses the fact that $\sin(\pi-x)=\sin x$ and may best be explained with the example

$$\begin{align}
{1\over16}\left({1\over\sin^2\left(\pi\over8\right)}+{1\over\sin^2\left(3\pi\over8\right)}+{1\over\sin^2\left(5\pi\over8\right)}+{1\over\sin^2\left(7\pi\over8\right)} \right)
&={1\over16}\left({1\over\sin^2\left(\pi\over8\right)}+{1\over\sin^2\left(3\pi\over8\right)}+{1\over\sin^2\left(3\pi\over8\right)}+{1\over\sin^2\left(\pi\over8\right)} \right)\\
&={2\over16}\left({1\over\sin^2\left(\pi\over8\right)}+{1\over\sin^2\left(3\pi\over8\right)} \right)

\end{align}$$