2016/019) If $\theta = \frac{\pi}{2^{n+1}}$ then show that $2^n cos\theta \ cos 2 \theta \cos 2^2\theta \cdots cos 2^n \theta = 1$

$2^n cos\theta \cos 2 \theta \cos 2^2\theta \cdots cos 2^n \theta$
$= (2\cos\theta) (2\cos 2 \theta)(2 \cos 2^2\theta) \cdots (2cos 2^n \theta)  = 1$
$= (\frac{\sin 2\theta}{\sin \theta}) (\frac{\sin 4\theta}{\sin 2\theta})
          \cdots (\frac{\sin 2^n\theta}{\sin 2^{n-1}\theta})  = 1$
$=\frac{\sin 2^n \theta }{\sin \theta}$
$==\frac{\sin 2^n\frac{\pi}{2^n+1})}{\sin \frac{\pi}{2^n+1}}$
$==\frac{\sin (\frac{\pi}{2^n}-\frac{\pi}{2^n+1}) }{\sin \frac{\pi}{2^n+1}}$
$==\frac{\sin (\frac{\pi}{2^n+1}) }{\sin \frac{\pi}{2^n+1}}=1 $