* Find all natural numbers $n$ and $k$ such that $n^n$ is a $k$-digit number, and $k^k$ is an $n$-digit number.
Parse $n^{n}$ is a $k$-digit number, $k^{k}$ is an $n$-digit number, i.e., $10^{k-1} \leqslant n^{n}<10^{k}, 10^{n-1} \leqslant k^{k}<10^{n}$. Without loss of generality, assume $n \geqslant k$, then $n^{n}<10^{n}$, i.e., $n<10, k<10$. Since $2^{2}<10,10<3^{3}<10^{2}, 10^{2}<$ $4^{4}<10^{3}, 10^{3}<5^{5}<10^{4}, 10^{4}<6^{6}<10^{5}, 10^{5}<7^{7}<10^{6}, 10^{7}<8^{8}<10^{8}$, $10^{8}<9^{9}<10^{9}$, so $n=k$, and $k=1,8$ or 9.