As $\sin 10^\circ = \cos 80^\circ$, $\sin 30^\circ = \cos 60^\circ$, $\sin 50^\circ = \cos 40^\circ$, and $\sin 70^\circ = \cos 20^\circ$, the desired equality is equivalent to
$$(\cos 20^\circ \cdot \cos 40^\circ \cdot \cos 60^\circ \cdot \cos 80^\circ)^2 = \frac{1}{256}.$$
It is known that $\cos 60^\circ = \frac{1}{2}$. Concerning the other factors, we obtain
$$\begin{aligned} \cos 20^\circ \cdot \cos 40^\circ \cdot \cos 80^\circ &= \frac{8 \sin 20^\circ \cos 20^\circ \cdot \cos 40^\circ \cdot \cos 80^\circ}{8 \sin 20^\circ} \\ &= \frac{4 \sin 40^\circ \cos 40^\circ \cdot \cos 80^\circ}{8 \sin 20^\circ} \\ &= \frac{2 \sin 80^\circ \cos 80^\circ}{8 \sin 20^\circ} \\ &= \frac{\sin 160^\circ}{8 \sin 20^\circ} = \frac{\sin 20^\circ}{8 \sin 20^\circ} = \frac{1}{8}. \end{aligned}$$
Consequently,
$$(\cos 20^\circ \cdot \cos 40^\circ \cdot \cos 60^\circ \cdot \cos 80^\circ)^2 = \left(\frac{1}{2} \cdot \frac{1}{8}\right)^2 = \left(\frac{1}{16}\right)^2 = \frac{1}{256}.$$

As $\sin 10^\circ = \cos 80^\circ$, $\sin 30^\circ = \cos 60^\circ$, $\sin 50^\circ = \cos 40^\circ$, and $\sin 70^\circ = \cos 20^\circ$, the l.h.s. of the desired equality can be rewritten as $(\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ)^2$. Hence the desired equality is equivalent to
$$\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{16}. \qquad (2)$$
As $\cos 60^\circ = \frac{1}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we have
$$\begin{aligned} \cos 40^\circ \cos 80^\circ &= \cos (60^\circ - 20^\circ) \cos (60^\circ + 20^\circ) \\ &= (\cos 60^\circ \cos 20^\circ + \sin 60^\circ \sin 20^\circ) (\cos 60^\circ \cos 20^\circ - \sin 60^\circ \sin 20^\circ) \\ &= \cos^2 60^\circ \cos^2 20^\circ - \sin^2 60^\circ \sin^2 20^\circ \\ &= \frac{1}{4} \cos^2 20^\circ - \frac{3}{4} \sin^2 20^\circ = \frac{1}{4} (\cos^2 20^\circ - 3 \sin^2 20^\circ). \end{aligned}$$
Thus
$$\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{8} (\cos^3 20^\circ - 3 \sin^2 20^\circ \cos 20^\circ). \quad (3)$$
By applying the formula $\cos 3x = \cos^3 x - 3 \sin^2 x \cos x$ to $x = 20^\circ$, we get $\cos^3 20^\circ - 3 \sin^2 20^\circ \cos 20^\circ = \cos 60^\circ = \frac{1}{2}$. Consequently, (3) reduces to (2), completing the solution of the problem.