$1 \cdot 29$ Find the largest positive integer $k$, such that
$$
1991^{k} \cdot 1990^{1991^{1992}}+1992^{1991^{1990}} \text {. }
$$
(32nd International Mathematical Olympiad Preliminary Problem, 1991)
[Solution] First, prove by mathematical induction:
Let $a \geqslant 3$ be an odd number, for all positive integers $n$, we have
$$
(1+a)^{a^{n}}=1+S_{n} a^{n+1} .
$$

where $S_{n}$ is an integer, and $a \nmid S_{n}$.
For $n=1$, we have
$$
\begin{aligned}
& (1+a)^{a} \\
= & 1+C_{a}^{1} a+C_{a}^{2} a^{2}+\cdots+C_{a}^{a} a^{a} \\
= & 1+a^{2}\left(1+C_{a}^{2}+C_{a}^{3} a+\cdots+a^{a-2}\right) .
\end{aligned}
$$

Since $a$ is odd, $a \mid C_{a}^{2}$, thus

therefore
$$
a \mid C_{a}^{2}+C_{a}^{3} a+\cdots+a^{a-2},
$$
$$
a \nmid S_{1}=1+C_{a}^{2}+\cdots+a^{a-2} \text {. }
$$

Therefore, for $n=1$, (1) holds.
Assume (1) holds for the natural number $n=k_{0}$, then
$$
\begin{aligned}
& (1+a)^{a^{k_{0}+1}} \\
= & {\left[(1+a)^{a^{k_{0}}}\right]^{a} } \\
= & \left(1+S_{k_{0}} a^{k_{0}+1}\right)^{a} . \\
= & 1+S_{k_{0}} a^{k_{0}+2}+C_{a}^{2} S_{k_{0}}^{2} a^{2 k_{0}+2}+\cdots+S_{k_{0}}^{a} a^{a\left(k_{0}+1\right)} \\
= & 1+S_{k_{0}+1} a^{k_{0}+2},
\end{aligned}
$$

where
$$
S_{k_{0}+1}=S_{k_{0}}+C_{a}^{2} S_{k_{0}}^{2} a^{k_{0}}+\cdots+S_{k_{0}}^{a} a^{a\left(k_{0}+1\right)-k_{0}-2} .
$$

By the induction hypothesis $a \nmid S_{k_{0}}$, thus $a \nmid S_{k_{0}+1}$.
Therefore, (1) holds for $n=k_{0}+1$.
That is, (1) holds for all natural numbers $n$.
Similarly, we can prove:
Let $b \geqslant 3$ be an odd number, then for all positive integers $n$, we have
$$
(b-1)^{b^{n}}=-1+T_{n} b^{n+1},
$$

where $T_{n}$ is an integer, and $b \nmid T_{n}$.
Using (1) and (2), we know that there exist integers $S$ and $T$, such that
$$
1991 \nmid S, 1991 \nmid T \text {, }
$$

and
$$
\begin{aligned}
& 1990^{1991^{1092}}+1992^{1991^{1900}} \\
= & T \cdot 1991^{1993}+S \cdot 1991^{1991} \\
= & 1991^{1991}\left(T \cdot 1991^{2}+S\right)
\end{aligned}
$$

Therefore, the largest $k=1991$.