When n=1, 2, \dots, n, we get the following equations by using double angle identity.

2\sin \dfrac{x}{2}\cos \dfrac{x}{2}=\sin x

2\sin \dfrac{x}{4}\cos \dfrac{x}{4} =\sin \dfrac{x}{2}

\dots

2\sin \dfrac{x}{2^n} \cdotp \cos \dfrac{x}{2^n}=\sin \dfrac{x}{2^{n-1}}

Multiply the above equations, cancel common factors

2^n\sin \dfrac{x}{2^n} \cos \dfrac{x}{2} \cos \dfrac{x}{4}\dots\cos \dfrac{x}{2^n}=\sin x

Divide two sides using 2^n\sin \dfrac{x}{2^n} to complete the proof

\cos \dfrac{x}{2}\cdotp \cos \dfrac{x}{4}\dots\cos \dfrac{x}{2^n}=\dfrac{\sin x}{2^n\sin \dfrac{x}{2^n} }