A Hidden Trigonometric Identity

From $\displaystyle\sin{2x}=2\sin{x}\cos{x}$ , $\sin{\frac{\pi}{4}}=4\sin{\frac{\pi}{16}}\cos{\frac{\pi}{16}}\cos{\frac{\pi}{8}}$ $=8\sin{\frac{\pi}{32}}\cos{\frac{\pi}{32}}\cos{\frac{\pi}{16}}\cos{\frac{\pi}{8}}$ $=16\sin{\frac{\pi}{64}}\cos{\frac{\pi}{64}}\cos{\frac{\pi}{32}}\cos{\frac{\pi}{16}}\cos{\frac{\pi}{8}}.$ Continually expanding the sine term results in the infinite product of cosines. Isolating the product, $\lim_{n\rightarrow\infty}\frac{\sin{\frac{\pi}{4}}}{2^n\sin{\frac{\pi}{4\cdot2^{n}}}}=\prod_{n=1}^{\infty}\cos{\frac{\pi}{4\cdot2^n}}.$ Applying L'Hopital's rule to the left side gives the desired solution.