Example 13 Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ all be natural numbers, and $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} x_{4} x_{5}$. Try to find the maximum value of $x_{5}$.
Solution: Without loss of generality, let $x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5}$.
$$
\because x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} \text {. }
$$
$x_{4} x_{5}$,
$$
\begin{aligned}
\therefore 1= & \frac{1}{x_{2} x_{3} x_{4} x_{5}}+\frac{1}{x_{1} x_{3} x_{4} x_{5}} \\
& +\frac{1}{x_{1} x_{2} x_{4} x_{5}}+\frac{1}{x_{1} x_{2} x_{3} x_{5}} \\
& +\frac{1}{x_{1} x_{2} x_{3} x_{4}} \\
\leqslant & \frac{1}{x_{4} x_{5}}+\frac{1}{x_{4} x_{5}}+\frac{1}{x_{4} x_{5}}+\frac{1}{x_{5}}+\frac{1}{x_{4}} \\
& =\frac{3+x_{4}+x_{5}}{x_{4} x_{5}} .
\end{aligned}
$$

Thus, $x_{4} x_{5} \leqslant 3+x_{4}+x_{5}$.
Therefore, $\left(x_{4}-1\right)\left(x_{5}-1\right) \leqslant 4$.
If $x_{4}=1$, then $x_{1}=x_{2}=x_{3}=x_{4}=1$. From the given, we have $4+x_{5}=x_{5}$, which is a contradiction. Therefore, $x_{4} \geqslant 2$. Then $x_{5}-1 \leqslant\left(x_{4}-1\right)\left(x_{5}-1\right) \leqslant 4, x_{5} \leqslant 5$.

When $x_{5}=5$, there exist $x_{1}=x_{2}=x_{3}=1$, $x_{4}=2$ that satisfy the equation.
Therefore, the maximum value of $x_{5}$ is 5.