1.39 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 $n \geqslant k$, then 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, by $3^{3}<100,4^{4}<10^{3}, 5^{5}<10^{4}, 6^{6}<10^{5}, 7^{7}<10^{6}$, we know that $n=3,4,5,6,7$ do not meet the requirement.

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

Therefore, the solutions to this problem are $n=k=1, n=k=8, n=k=9$.