10.33 Find all natural numbers $n, k$ such that $n^{n}$ has $k$ digits, and $k^{k}$ has $n$ digits.
(8th All-Soviet Union Mathematical Olympiad, 1974)
[Solution] For definiteness, let $n \geqslant k$.
If $n^{n}$ has $k$ digits, then
$$
10^{k-1} \leqslant n^{n}<10^{k} .
$$

If $k^{k}$ has $n$ digits, then
$$
10^{n-1} \leqslant k^{k}<10^{n} \text {. }
$$

From $n \geqslant k$ and (1), we get
$$
n^{n}<10^{k} \leqslant 10^{n} .
$$

Thus, $n<10$. Therefore, $k<10$.
For $k=1,2,3,4,5,6,7,8,9$, we verify one by one:
$$
\begin{array}{l}
10^{0} \leqslant 1<10^{1}, \quad 2^{2}<10^{1}, \quad 3^{3}<10^{2}, \\
4^{4}<10^{3}, \quad 5^{5}<10^{4}, \quad 6^{6}<10^{5}, \quad 7^{7}<10^{6}, \\
10^{7}<8^{8}<10^{8}, \quad 10^{8}<9^{9}<10^{9},
\end{array}
$$

From (1), (2), and the above verification, we get
$$
n=k=1,8 \text { or } 9 \text {. }
$$