$8 \cdot 15$ Let $a_{1}=3, b_{1}=100$, for $n \geqslant 1$
$$a_{n+1}=3^{a_{n}}, b_{n+1}=100^{b}{ }_{n}$$

Find the smallest positive integer $m$ such that $b_{m}>a_{100}$.
[Solution] Since $3^{3}a_{100}$, we have $m \leqslant 99$.
On the other hand, we can prove by induction that
$$a_{n+2}>6 b_{n}, n \geqslant 1$$

In fact, when $n=1$, we have $a_{3}=3^{27}>6 b_{1}=600$, so (1) holds. Suppose (1) holds for $n=k$. When $n=k+1$, we have
$$a_{k+3}=3^{a_{k+2}}>3^{6 b_{k}}>6 \cdot 100^{b_{k}}=6 b_{k+1}.$$

Thus, (1) also holds. Therefore, (1) holds for any natural number $n$.
From (1), we get $b_{98}98$. In conclusion,
$$m=99.$$