$1 \cdot 39$ Find all natural numbers $n, k$ such that $n^{n}$ has $k$ digits, and $k^{k}$ has $n$ digits.
[Solution] $n=k$, and $k=1, 8$ or 9.
If $n^{n}$ has $k$ digits, and $k^{k}$ has $n$ digits, then
$$
\begin{array}{l}
10^{k-1}<n^{n}<10^{k} \\
10^{n-1}<k^{k}<10^{n}
\end{array}
$$

Assume without loss of generality that $n \geqslant k$, thus we have
$$
n<10 \text { and } k<10 \text {. }
$$

If $n=1$, then since $n^{n}$ has one digit, $k=1$, and it is verified that $n=k=1$ is a solution to the problem.

If $n=2$, then since $n^{n}=4$ has one digit, $k=1$, but in this case $k^{k}$ does not have $n$ digits, which does not meet the requirement.

Similarly, from $3^{3}<100, 4^{4}<10^{3}, 5^{5}<10^{4}, 6^{6}<10^{5}, 7^{7}<10^{6}$, we know that $n=3,4,5,6,7$ do not meet the requirement.

From $10^{7}<8^{8}<10^{8}, 10^{8}<9^{9}<10^{9}$, we know that $n=k=8, n=k=9$ are solutions to the problem.
Therefore, the solutions to this problem are $n=k=1, n=k=8, n=k=9$.