This is a typical problem that can be solved by the telescoping method. \begin{align} &\cos{x}\cos{2x}\cdots\cos{2^{n-1}x}\\ &=\frac{\sin{2x}}{2\sin{x}}\cdot\frac{\sin{4x}}{2\sin{2x}}\cdots\frac{\sin{2^nx}}{2\sin{2^{n-1}x}}\\ &=\boxed{\frac{\sin{2^nx}}{2^n\sin{x}}} \end{align}