Example 9 Divide 1998 into 5 positive integers $a_{i}(i=1,2, \cdots, 5)$, so that $a_{1}!\cdot a_{2}!\cdot a_{3}!\cdot a_{4}!\cdot a_{5}!$ has the minimum value.
For $m, n \in \mathbf{N}$, if $m > n + 1$, then we have
$$
\begin{aligned}
m! \cdot n! & = (m-1)! \cdot m \cdot n! \\
& > (m-1)!(n+1)!
\end{aligned}
$$

This tells us that to minimize the product under discussion, the $a_i$ must be as close as possible.

Since $1998 = 399 \times 5 + 3$, the minimum value sought is $(399!)^2(400!)^3$.