63. Let $a_{1}=1997^{1997^{1997}}{ }^{.197}$ (1997 sevens in total), the sum of the digits in the decimal representation of $a_{1}$ is $a_{2}$, the sum of the digits of $a_{2}$ is $a_{3}$, and so on. Find $a_{2000}$.
Solution: Let $x_{n}=10000^{10000^{10000}}$ (with $n$ 10000s), $n=1,2, \cdots$, then for $n \geqslant 2$, we have
$$
\begin{array}{l} 
x_{n}=10000^{x_{n-1}}=10^{4 x_{n-1}} . \\
\therefore a_{1}<x_{1997}=10^{4 x_{1996}}, \\
a_{2} \leqslant 9 \times 4 x_{1996}<100 x_{1996} \\
=100 \cdot 10^{4 x 1995}=10^{4 x_{1995}+2}, \\
a_{3} \leqslant 9\left(4 x_{1995}+2\right)<9 \times 6 x_{1995}<100 x_{1995} \\
=10^{4 x_{1994}+2},
\end{array}
$$

Thus, for any $k$, when $1 \leqslant k \leqslant 1996$, we have
$$
a_{k}<10^{4 x} 1997-k{ }^{+2} \text {. }
$$

When $k=1996$, i.e.,
$$
\begin{array}{l}
a_{1996}<10^{4 x_{1}+2}, \\
\therefore a_{1997}<9 \times\left(4 x_{1}+2\right)=9 \times 40002<10^{6}, \\
a_{1998}<9 \times 6=54, \\
a_{1999} \leqslant 13, \\
a_{2000} \leqslant 9 .
\end{array}
$$

Since $a_{2000} \geqslant 1$,
$$
\therefore 1 \leqslant a_{20000} \leqslant 9 \text {. }
$$

From the generation of $a_{2}, a_{3}, \cdots, a_{2000}$, we know that
$$
a_{1} \equiv a_{2} \equiv a_{3} \equiv \cdots \equiv a_{2000}(\bmod 9) \text {. }
$$

It is easy to see that $1997=9 m-1(m=222)$, and the exponent of $a_{1}$ is odd, denoted as $r$.
$$
\begin{aligned}
\therefore a_{1} & =1997^{r}=(9 m-1)^{r} \\
& =9 s+(-1)^{r}=9 s-1, \\
a_{2000} & =a_{1}=-1(\bmod 9) .
\end{aligned}
$$

Therefore, $a_{23000}=8$.
(Qin Zongci, Jiangsu Zhenjiang Teachers College, 212003)