I love cos

More simple way to solve.. $\text{assume that } \frac{\pi}{7}= \theta$ $7\theta = \pi$ $\text{so} \frac{1}{2sin\theta} 2sin\theta cos\theta cos2\theta cos (7\theta - 4\theta)$ $\frac{1}{2sin\theta} sin2\theta cos2\theta cos (\pi - 4\theta )$ $\frac{-1}{4sin\theta} sin4\theta cos4\theta = \frac{-1}{8sin\theta} sin8\theta$ $\frac{-1}{8sin\theta} (\pi +\theta)= \frac{1}{8sin\theta} sin\theta = \frac{1}{8}$

$\large{\displaystyle \prod _{ k=1 }^{ n-1 }{ \cos { \left( \frac { k\pi }{ n } \right) } =\frac { \sin { \left( \frac { n\pi }{ 2 } \right) } }{ { 2 }^{ n-1 } } } }$

using $\cos(x)=-\cos (\pi-x)$ and simplify the expression.

$\huge{-\left( \cos \left(\frac{\pi}{7} \right)\cos \left(\frac{2 \pi}{7} \right) \cos \left(\frac{3 \pi}{7} \right) \right)^{2}=\frac{\sin \left(\frac{7 \pi}{2} \right)}{2^{n-1}}}$

$\large{\left( \cos \left(\frac{\pi}{7} \right)\cos \left(\frac{2 \pi}{7} \right) \cos \left(\frac{3 \pi}{7} \right) \right)^{2}=\frac{1}{2^{6}}}$

As angles are acute so answer is positive