$8 \cdot 2$ The sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1989^{1989}$, for $n>1, a_{n}$ is equal to the sum of the digits of $a_{n-1}$. Try to find $a_{5}$.
[Solution] Since $a_{1}=1989^{1989}<10^{4 \cdot 1989}$,
then $a_{2} \leqslant 9 \cdot(4 \cdot 1989+1)<80000$,
which means $a_{2}$ has at most 5 digits. Therefore,
$$
a_{3} \leqslant 45, \quad a_{4} \leqslant 12, \quad a_{5} \leqslant 9 .
$$

Since $9 \mid 1989$, it follows that $9 \mid a_{5}$, and obviously $a_{5} \neq 0$, thus $a_{5}=9$.