Yet another curious identity?

Angles will be measured in degrees throughout.

Multiplying the given equation with $\sin(8)$ and using the rule $2(\sin{x})(\cos{x})$ $=\sin(2x)$ repeatedly, we find that $\sin(2^{n+1})=\sin(8).$ (This approach is discussed here .)This means that either $2^{n+1}-8=360k$ or $2^{n+1}+8=180(2k-1)$ for some positive integer $k$ . The second equation has no integer solutions since the LHS is divisible by 8 while the RHS isn't. Dividing the first equation by 8, we see that $2^{n-2}-1=45k$ , meaning that $2^{n-2}\equiv{1}\pmod{45}$ . Since the Carmichael lambda of 45 is 12, we know that $2^{12}\equiv{1}\pmod{45}$ , and we can easily check that no lower exponent does the job. Thus $n-2=12$ and $n=\boxed{14}$ . In summary, we have the intriguing "Morrie's Law" type of equation $(\cos{8})(\cos{16})(\cos{32})...(\cos{8192})(\cos{16384})=\prod_{k=3}^{14}\cos(2^k)=\frac{1}{2^{12}}$

Now, what about the bonus question?