Example 4. Try to decompose 1988 into 8 positive integers $a_{1}$, $a_{2}$, ..., $a_{8}$, so that $a_{1}!a_{4}! \cdots a_{8}!$ is minimized.
$$
\begin{array}{l}
\quad \text { Solving } 1988=8 \times 248+4 . \\
\quad \text { Let } a_{1}=a_{2}=a_{3}=a_{4}=248, a_{5}=a_{6}=a_{7} \\
=a_{8}=249 \text {. }
\end{array}
$$

This makes $a_{1}!\cdot a_{2}!\cdot \cdots \cdot a_{8}!$ the smallest.
