Let's calculate the value of the following integral:

$$
\int_{0}^{\pi / 2^{n+1}} \sin x \cdot \cos x \cdot \cos 2 x \cdot \cos 2^{2} x \cdot \ldots \cdot \cos 2^{n-1} x d x
$$
The integrand product can be transformed into the product of a single trigonometric function and a constant by applying the identity

$$
\sin u \cos u=\frac{1}{2} \sin 2 u
$$

$n$ times, with $u$ successively taking the values $x$, $2x$, $2^2 x$, ..., $2^{n-1} x$. Thus, the consecutive cosine factors are always preceded by, or appear as, the sine of the same argument, so

$$
I=\int_{0}^{a} \frac{1}{2^{n}} \sin 2^{n} x d x, \quad \text { where } \quad a=\frac{\pi}{2^{n+1}}
$$

Finally, we transform the constant into a product such that one of its factors is equal to the derivative of the inner function of the composite function $\sin 2^{n} x$, thus

$$
I=\frac{1}{2^{2 n}} \int_{0}^{a} 2^{n} \sin 2^{n} x d x=\frac{1}{2^{2 n}}\left[-\cos 2^{n} x\right]_{0}^{a}
$$

and since $2^{n} a=\pi / 2$, therefore $I=1 / 2^{2 n}$.