Prove that $\frac{\sin x}{x}=(\cos\frac{x}{2}) (\cos\frac{x}{4}) (\cos \frac{x}{8})...$ How do I prove this identity:
$$\frac{\sin x}{x}=\left(\cos\frac{x}{2}\right) \left(\cos\frac{x}{4}\right) \left(\cos \frac{x}{8}\right)...$$
My idea is to let 
$$y=\frac{\sin x}{x}$$
and
$$xy=\sin x$$
Then use the double angle identity $\sin 2x=2\sin x \cos x$ and its half angle counterparts repeatedly.  I see some kind of pattern, but I can't seem to make out the pattern and complete the proof.

Note the fact that
$$
\cos \frac{x}{2^k} = \frac12 \cdot \frac{\sin (2^{1-k} x)}{\sin(2^{-k}x)},
$$
and we have
$$
\prod_{k = 1}^n \cos \frac{x}{2^k} = \frac{1}{2^n} \cdot \frac{\sin x}{\sin(2^{-n}x)} = \frac{2^{-n}x}{\sin(2^{-n}x)} \cdot \frac{\sin x}{x}.
$$
For all $x$, as $n \to \infty$, we have
$$
\lim_{n \to \infty} \prod_{k = 1}^n \cos \frac{x}{2^k}=  \frac{\sin x}{x} \cdot\lim_{n \to \infty} \frac{2^{-n}x}{\sin(2^{-n}x)} = \frac{\sin x}{x}.
$$
