Example 2 Find
$$
I=\sin \frac{\pi}{n} \cdot \sin \frac{2 \pi}{n} \cdots \cdots \sin \frac{(n-1) \pi}{n}
$$

the value of ( $n$ is a natural number greater than 1).
Let $\omega=\cos \frac{\pi}{n}+\mathrm{i} \sin \frac{\pi}{n}$. Then $\omega^{2 n}=1$.
Thus, $1, \omega^{2}, \omega^{4}, \cdots, \omega^{2(n-1)}$ are all roots of $x^{2 n}-1=0$.
Therefore, $x^{2 n}-1$
$$
\begin{array}{l}
=\left(x^{2}-1\right)\left[x^{2(n-1)}+x^{2(n-2)}+\cdots+x^{2}+1\right] \\
=\left(x^{2}-1\right)\left(x^{2}-\omega^{2}\right)\left(x^{2}-\omega^{4}\right) \cdots\left[x^{2}-\omega^{2(n-1)}\right] .
\end{array}
$$

Hence, $x^{2(n-1)}+x^{2(n-2)}+\cdots+x^{2}+1$
$$
=\left(x^{2}-\omega^{2}\right)\left(x^{2}-\omega^{4}\right) \cdots\left[x^{2}-\omega^{2(n-1)}\right] \text {. }
$$

Let $x=1$, we get
$$
\left(1-\omega^{2}\right)\left(1-\omega^{4}\right) \cdots\left[1-\omega^{2(n-1)}\right]=n \text {. }
$$

And $\sin \frac{k \pi}{n}=\frac{\omega^{k}-\omega^{-k}}{2 \mathrm{i}}=\frac{\omega^{2 k}-1}{2 \mathrm{i} \cdot \omega^{k}}$
$$
\begin{aligned}
\Rightarrow I & =\sin \frac{\pi}{n} \cdot \sin \frac{2 \pi}{n} \cdots \cdots \cdot \sin \frac{(n-1) \pi}{n} \\
& =\frac{\left(\omega^{2}-1\right)\left(\omega^{4}-1\right) \cdots\left[\omega^{2(n-1)}-1\right]}{(2 \mathrm{i})^{n-1} \omega^{\frac{1}{2} n(n-1)}} .
\end{aligned}
$$

Thus, $(2 \mathrm{i})^{n-1} \omega^{\frac{1}{2} n(n-1)}=(2 \mathrm{i})^{n-1}\left(\omega^{\frac{n}{2}}\right)^{n-1}$
$$
\begin{array}{l}
=2^{n-1}(-1)^{n-1} \\
\Rightarrow I=\frac{\left(\omega^{2}-1\right)\left(\omega^{4}-1\right) \cdots\left[\omega^{2(n-1)}-1\right]}{2^{n-1}(-1)^{n-1}} \\
=\frac{\left(1-\omega^{2}\right)\left(1-\omega^{4}\right) \cdots\left[1-\omega^{2(n-1)}\right]}{2^{n-1}} \\
=\frac{n}{2^{n-1}} .
\end{array}
$$