Example 3 Calculate $\tan \frac{\pi}{7} \cdot \tan \frac{2 \pi}{7} \cdot \tan \frac{3 \pi}{7}$. (2016, Harvard-MIT Mathematics Tournament)
【Analysis】Consider the polynomial $P(z)=z^{7}-1$.
Let $z=\mathrm{e}^{\mathrm{i} x}=\cos x+\mathrm{i} \sin x$.
Then $z^{7}-1$
$$
\begin{aligned}
= & \left(\cos ^{7} x-\mathrm{C}_{7}^{2} \cos ^{5} x \cdot \sin ^{2} x+\right. \\
& \left.\mathrm{C}_{7}^{4} \cos ^{3} x \cdot \sin ^{4} x-\mathrm{C}_{7}^{6} \cos x \cdot \sin ^{6} x-1\right)+ \\
& \mathrm{i}\left(-\sin ^{7} x+\mathrm{C}_{7}^{2} \sin ^{5} x \cdot \cos ^{2} x-\right. \\
& \left.\mathrm{C}_{7}^{4} \sin ^{3} x \cdot \cos ^{4} x+\mathrm{C}_{7}^{6} \sin x \cdot \cos ^{6} x\right) .
\end{aligned}
$$

First, consider the real part of the above expression.
After simplification, the roots of $64 \cos ^{7} x-\cdots-1=0$ are $x=\frac{2 \pi}{7}, \frac{4 \pi}{7}, \cdots, 2 \pi$.
By Vieta's formulas, $\prod_{k=1}^{7} \cos \frac{2 k \pi}{7}=\frac{1}{64}$.
Also, $\prod_{k=1}^{7} \cos \frac{2 k \pi}{7}=\left(\prod_{k=1}^{3} \cos \frac{k \pi}{7}\right)^{2}$, hence
$$
\prod_{k=1}^{3} \cos \frac{k \pi}{7}=\frac{1}{8} \text {. }
$$

Next, consider the imaginary part.
After simplification, we have
$$
\begin{array}{l}
-64 \sin ^{7} x+\cdots+7 \sin x \\
=\left(-64 \sin ^{6} x+\cdots+7\right) \sin x .
\end{array}
$$

Then the roots of $-64 \sin ^{6} x+\cdots+7=0$ are $x=\frac{2 \pi}{7}, \frac{4 \pi}{7}, \cdots, \frac{12 \pi}{7}$.
Thus, $\prod_{k=1}^{6} \sin \frac{2 k \pi}{7}=-\frac{7}{64}$.
Similarly, $\prod_{k=1}^{6} \sin \frac{2 k \pi}{7}=-\left(\prod_{k=1}^{3} \sin \frac{2 k \pi}{7}\right)^{2}$. Hence $\prod_{k=1}^{6} \sin \frac{2 k \pi}{7}=-\left(\prod_{k=1}^{3} \sin \frac{k \pi}{7}\right)^{2}$, $\prod_{k=1}^{3} \sin \frac{k \pi}{7}=\frac{\sqrt{7}}{8}$.
Therefore, $\tan \frac{\pi}{7} \cdot \tan \frac{2 \pi}{7} \cdot \tan \frac{3 \pi}{7}=\sqrt{7}$.
【Note】Example 3 has more general conclusions:
(1) $\prod_{k=1}^{n} \cos \frac{k \pi}{2 n+1}=\frac{1}{2^{n}}$;
(2) $\prod_{k=1}^{n} \sin \frac{k \pi}{2 n+1}=\frac{\sqrt{2 n+1}}{2^{n}}$;
(3) $\prod_{k=1}^{n} \tan \frac{k \pi}{2 n+1}=\sqrt{2 n+1}$.