7. $\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{3 \pi}{15} \cdots \cos \frac{7 \pi}{15}$ The value is $\qquad$
7. $\frac{1}{128}$ Using the formula $\cos \theta \cdot \cos \left(\frac{\pi}{3}-\theta\right) \cdot \cos \left(\frac{\pi}{3}+\theta\right)=\frac{1}{4} \cos 3 \theta$, we get:
$$
\begin{array}{l}
\cos \frac{\pi}{15} \cdot \cos \left(\frac{\pi}{3}-\frac{\pi}{15}\right) \cdot \cos \left(\frac{\pi}{3}+\frac{\pi}{15}\right)=\cos \frac{\pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{6 \pi}{15}=\frac{1}{4} \cos \frac{\pi}{5} \\
\cos \frac{2 \pi}{15} \cdot \cos \left(\frac{\pi}{3}-\frac{2 \pi}{15}\right) \cdot \cos \left(\frac{\pi}{3}+\frac{2 \pi}{15}\right)=\cos \frac{2 \pi}{15} \cdot \cos \frac{3 \pi}{15} \cdot \cos \frac{7 \pi}{15}=\frac{1}{4} \cos \frac{2 \pi}{5}
\end{array}
$$

And $\cos \frac{\pi}{5} \cos \frac{2 \pi}{5}=\frac{\sin \frac{4 \pi}{5}}{4 \sin \frac{\pi}{5}}=\frac{1}{4}$
So the original expression $=\frac{1}{16} \times \frac{1}{4} \times \frac{1}{2}=\frac{1}{128}$