If sinx cannot = 0, use mathematical induction to show that

$\displaystyle cosx \cdot cos2x \cdot cos4x... cos2^{n-1} x = \frac{sin2^n x}{2^n sinx}$ for every integer $\displaystyle n \geq 1$

So for the induction step what I have so far is:

Assume n=k is true, show true for n=k+1

$\displaystyle cosx \cdot cos2x \cdot cos4x... cos2^{k-1} x \cdot cos2^{(k+1)-1}x = \frac{sin2^{k+1} x}{2^{k+1} sinx}$

$\displaystyle \frac{sin2^{k}x}{2^{k}sinx} \cdot cos2^{k}x = \frac{sin2^{k+1}x}{2^{k+1}sinx}$

$\displaystyle \frac{(sin2^{k}x)(cos2^{k}x)}{2^{k}sinx} = \frac{sin2^{k+1}x}{2^{k+1}sinx}$

Assuming any of this is correct so far, I am having trouble showing that both sides are equal from here on.