Example 4 Solve the equation $w!=x!+y!+z!$. (1983, Canadian High School Mathematics Competition)
Solution: Without loss of generality, let $x \leqslant y \leqslant z$.
Obviously, $w \geqslant z+1$, then
$(z+1)! \leqslant w! = x! + y! + z!$
$\leqslant 3 \cdot z!$.
Thus, $(z+1) \leqslant 3$, i.e., $z \leqslant 2$.
When $z=0$ or $z=1$, the original equation has no solution. When $z=2$, the solution to the original equation is
$$
x=y=z=2, w=3 \text{.}
$$

Note. In some mathematical problems, estimating the range of variables often helps to solve the problem smoothly. Therefore, pay attention to this strategy when solving problems.