Concept:
Use the identity \(\cos(\pi - \theta) = -\cos\theta\) to pair terms, then apply the double-angle formula repeatedly to evaluate the product of cosines.
Step 1: Pair the cosine terms using symmetry.
Let
\[
P = \cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7}\cos\frac{4\pi}{7}\cos\frac{5\pi}{7}\cos\frac{6\pi}{7}.
\]
Note \(\cos\frac{6\pi}{7} = -\cos\frac{\pi}{7}\), \(\cos\frac{5\pi}{7} = -\cos\frac{2\pi}{7}\), \(\cos\frac{4\pi}{7} = -\cos\frac{3\pi}{7}\). Therefore,
\[
P = \left(-\cos\frac{\pi}{7}\right)\left(-\cos\frac{2\pi}{7}\right)\left(-\cos\frac{3\pi}{7}\right) \cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7} = -\left(\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7}\right)^2.
\]
Step 2: Evaluate \(\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7}\).
Let \(x = \frac{\pi}{7}\). Multiply and divide by \(2\sin x\):
\[
\cos x \cos 2x \cos 3x = \frac{2\sin x \cos x \cos 2x \cos 3x}{2\sin x} = \frac{\sin 2x \cos 2x \cos 3x}{2\sin x} = \frac{\sin 4x \cos 3x}{4\sin x}.
\]
Since \(\sin 4x = \sin\left(\frac{4\pi}{7}\right) = \sin\left(\pi - \frac{3\pi}{7}\right) = \sin 3x\), we get:
\[
\frac{\sin 3x \cos 3x}{4\sin x} = \frac{\sin 6x}{8\sin x}.
\]
Now \(\sin 6x = \sin\frac{6\pi}{7} = \sin\left(\pi - \frac{\pi}{7}\right) = \sin x\). Thus,
\[
\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7} = \frac{\sin x}{8\sin x} = \frac{1}{8}.
\]
Step 3: Substitute back and write the final answer.
\[
P = -\left(\frac{1}{8}\right)^2 = -\frac{1}{64}.
\]
\[
\boxed{-\frac{1}{64}}
\]