Example 1 Given that $p$ is a prime number greater than 3. Find
$$
\prod_{k=1}^{p-1}\left(1+2 \cos \frac{2 k \pi}{p}\right)
$$

the value.
Let $\omega=\mathrm{e}^{\frac{2 \pi i}{p}}$. Then
$$
\begin{array}{l} 
\omega^{p}=1, \omega^{-\frac{p}{2}}=-1, 2 \cos \frac{2 k \pi}{p}=\omega^{k}+\omega^{-k} . \\
\text { Hence } \prod_{k=1}^{p-1}\left(1+2 \cos \frac{2 k \pi}{p}\right)=\prod_{k=1}^{p-1}\left(1+\omega^{k}+\omega^{-k}\right) \\
=\prod_{k=1}^{p-1} \omega^{-k}\left(\omega^{2 k}+\omega^{k}+1\right) \\
=\omega^{-\left(\frac{p(p-1)}{2}\right)} \prod_{k=1}^{p-1}\left(\omega^{2 k}+\omega^{k}+1\right) \\
=(-1)^{p-1} \prod_{k=1}^{p-1} \frac{1-\omega^{3 k}}{1-\omega^{k}} .
\end{array}
$$

Since $p$ is a prime number greater than 3, $(-1)^{p-1}=1$; also, $p$ and 3 are coprime, $\{1,2, \cdots, p-1\}$ and $\{3 \times 1$, $3 \times 2, \cdots, 3 \times(p-1)\}$ both represent the reduced residue system modulo $p$.
Thus, $\prod_{k=1}^{p-1}\left(1-\omega^{3 k}\right)=\prod_{k=1}^{p-1}\left(1-\omega^{k}\right)$.
In conclusion, $\prod_{k=1}^{p-1}\left(1+2 \cos \frac{2 k \pi}{p}\right)=1$.