Given $n(n \geqslant 2)$ real numbers $\theta_{1}, \theta_{2}, \cdots, \theta_{n}\left(\theta_{i} \in\left(0, \frac{\pi}{2}\right], 1 \leqslant i \leqslant n\right)$ satisfying $\sum_{i=1}^{n} \sin \theta_{i} \leqslant 1$. Try to find the maximum value of $\prod_{i=1}^{n} \tan \theta_{i}$.
Let $\theta_{1}=\theta_{2}=\cdots=\theta_{n}=\arcsin \frac{1}{n}$, then
$$
\begin{array}{l}
\tan \theta_{1}=\tan \theta_{2}=\cdots=\tan \theta_{n}=\sqrt{\frac{1}{n^{2}-1}}, \\
\prod_{i=1}^{n} \tan \theta_{i}=\left(n^{2}-1\right)^{-\frac{n}{2}}
\end{array}
$$

We need to prove
$$
\prod_{i=1}^{n} \tan \theta_{i} \leqslant\left(n^{2}-1\right)^{-\frac{n}{2}}
$$

which is equivalent to
$$
\prod_{i=1}^{n} \cos ^{2} \theta_{i} \geqslant\left(n^{2}-1\right)^{n} \prod_{i=1}^{n} \sin ^{2} \theta_{i}
$$

or
$$
\prod_{i=1}^{n} \frac{1-\sin ^{2} \theta_{i}}{\sin ^{2} \theta_{i}} \geqslant\left(n^{2}-1\right)^{n} \Leftrightarrow \prod_{i=1}^{n} \frac{\left(1+\sin \theta_{i}\right)\left(1-\sin \theta_{i}\right)}{\sin ^{2} \theta_{i}} \geqslant\left(n^{2}-1\right)^{n} .
$$

Thus, it suffices to prove
$$
\prod_{i=1}^{n} \frac{\left(\sin \theta_{1}+\sin \theta_{2}+\cdots+\sin \theta_{n}+\sin \theta_{i}\right)\left(\sin \theta_{1}+\sin \theta_{2}+\cdots+\sin \theta_{n}-\sin \theta_{i}\right)}{\sin ^{2} \theta_{i}} \geqslant\left(n^{2}-1\right)^{n},
$$

which is
$$
\begin{array}{l}
\prod_{i=1}^{n} \frac{\left(\sin \theta_{1}+\sin \theta_{2}+\cdots+\sin \theta_{n}+\sin \theta_{i}\right)\left(\sin \theta_{1}+\cdots+\sin \theta_{i-1}+\sin \theta_{i+1}+\cdots+\sin \theta_{n}\right)}{\sin ^{2} \theta_{i}} \\
\geqslant\left(n^{2}-1\right)^{n}
\end{array}
$$

By the AM-GM inequality, we have
$$
\begin{array}{l}
\prod_{i=1}^{n} \frac{\left(\sin \theta_{1}+\sin \theta_{2}+\cdots+\sin \theta_{n}+\sin \theta_{i}\right)\left(\sin \theta_{1}+\cdots+\sin \theta_{i-1}+\sin \theta_{i+1}+\cdots+\sin \theta_{n}\right)}{\sin ^{2} \theta_{i}} \\
\geqslant \prod_{i=1}^{n} \frac{(n+1)\left(\sin \theta_{1} \sin \theta_{2} \cdots \sin \theta_{n} \sin \theta_{i}\right)^{\frac{1}{n+1}} \cdot(n-1)\left(\sin \theta_{1} \cdots \sin \theta_{i-1} \sin \theta_{i+1} \cdots \sin \theta_{n}\right)^{\frac{1}{n-1}}}{\sin ^{2} \theta_{i}} \\
=\left(n^{2}-1\right)^{n} \cdot \frac{\left(\left(\sin \theta_{1} \sin \theta_{2} \cdots \sin \theta_{n}\right)^{n}\left(\sin \theta_{1} \sin \theta_{2} \cdots \sin \theta_{n}\right)\right)^{\frac{1}{n+1}}\left(\frac{\left(\sin \theta_{1} \sin \theta_{2} \cdots \sin \theta_{n}\right)^{n}}{\sin \theta_{1} \sin \theta_{2} \cdots \sin \theta_{n}}\right)^{n-1}}{\prod_{i=1}^{n} \sin ^{2} \theta_{i}} \\
=\left(n^{2}-1\right)^{n} .
\end{array}
$$

In conclusion, the maximum value of $\prod_{i=1}^{n} \tan \theta_{i}$ is $\left(n^{2}-1\right)^{-\frac{n}{2}}$.