Prove: $ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $ 
Prove 
  $$ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $$
  where $n \in \mathbb{N}$ and $x$ not a multiple of $\frac{ \pi }{2^k} $ for any $k \in \mathbb{N}$.

My try. If $n=2$, we have
$$\begin{align}
\frac{1}{\sin 2x} + \frac{ 1}{\sin 4 x} &= \frac{1}{\sin 2x} + \frac{1}{2 \sin 2x \cos 2x } \\[6pt]
&= \frac{ 2 \cos 2x + 1 }{2 \sin 2x \cos 2x} \\[6pt]
&= \frac{2 \cos^2 x - 2 \sin^2 x + \cos^2 x + \sin^2 x}{2 \sin 2x \cos 2x} \\[6pt]
&= \frac{3 \cos^2 x - \sin^2 x}{2 \sin 2x \cos 2x}
\end{align}$$
but here I got stuck. I am on the right track? My goal is to ultimately use induction.

Let $S_n$ be the sum $\sum_{k=1}^n\frac{1}{\sin(2^kx)}$.
we want prove by induction that
$\forall n\geq1 \;\; S_n=cot(x)-cot(2^nx)$.
the formula is true for $n=1$.
let $n\geq1 : S_n true$.
we must prove that
$S_{n+1}-S_n=cot(2^nx)-cot(2^{n+1})$.
$cot(2^nx)-cot(2^{n+1}x)=$
$\frac{\cos(2^nx)}{\sin(2^nx)}-\frac{\cos(2^{n+1}x)}{2\sin(2^nx)\cos(2^nx)}=$
$\frac{1}{\sin(2^{n+1})}=S_{n+1}-S_n$.
using $\sin(2X)=2\sin(X)\cos(X)$
and
$\cos(2X)=2\cos^2(X)-1$.
thus
$\color{green}{S_{n+1}=S_n+cot(2^nx)-cot(2^{n+1})
=cot(x)-cot(2^{n+1})}$.
