Example 21 Let $x_{1}, x_{2}, \cdots, x_{n}$ all be natural numbers, and satisfy $x_{1}+x_{2}+\cdots+x_{n}=x_{1} x_{2} \cdots x_{n}$.
Find the maximum value of $x_{1}, x_{2}, \cdots, x_{n}$ $(n \geqslant 2)$.
By the symmetry of the conditions, without loss of generality, assume
$$
x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n} \text {. }
$$

This changes the symmetry of the conditions, which is equivalent to adding a condition
$$
x_{n-1} \geqslant 2 \quad (n \geqslant 2) \text {. }
$$

Otherwise, if $x_{n-1}=1$, from (2) we have
$$
x_{1}=x_{2}=\cdots=x_{n-2}=x_{n-1}=1 \text {, }
$$

Thus, substituting into (1) gives $(n-1)+x_{n}=x_{n}$, which is a contradiction. At this point, from (1) we have
$$
\begin{aligned}
x_{n} & =\frac{x_{1}+x_{2}+\cdots+x_{n-1}}{x_{1} x_{2} \cdots x_{n-1}-1} \\
& \leqslant \frac{x_{1} x_{2} \cdots x_{n-2}+\cdots+x_{1} x_{2} \cdots x_{n-2}+x_{1} x_{2} \cdots x_{n-2} x_{n-1}}{x_{1} x_{2} \cdots x_{n-1}-1} \\
& =\frac{\left(n-2+x_{n-1}\right) x_{1} x_{2} \cdots x_{n-2}}{x_{1} x_{2} \cdots x_{n-1}-1} \\
& \leqslant \frac{\left(n-2+x_{n-1}\right) x_{1} x_{2} \cdots x_{n-2}}{x_{1} x_{2} \cdots x_{n-1}-x_{1} x_{2} \cdots x_{n-2}} \\
& =\frac{n-2+x_{n-1}}{x_{n}-1} \\
& =1+\frac{n-1}{x_{n-1}-1} \leqslant n .
\end{aligned}
$$

When $x_{1}=x_{2}=\cdots=x_{n-2}=1$ and $x_{n-1}=2$, $x_{n}$ reaches its maximum value $n$. This is also the maximum value of $x_{1}, x_{2}, \cdots, x_{n}$.