Prove that $\frac{\sin x}{x}=(\cos\frac{x}{2}) (\cos\frac{x}{4}) (\cos \frac{x}{8})...$

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 $$ \lim_{n \to \infty} \prod_{k = 1}^n \cos \frac{x}{2^k}= \frac{\sin x}{x} \cdot\lim_{n \to \infty} \frac{2^{-n}x}{\sin(2^{-n}x)} = \frac{\sin x}{x}. $$