18. For integers $a>0, n>0$ and $\alpha \geq 0$, we shall write $a^{\alpha} \| n$ when $a^{\alpha} \mid n$ and $a^{\alpha+1} \nmid n$.
Lemma. For every odd number $a \geq 3$ and an integer $n \geq 0$ it holds that
$$
a^{n+1} \|(\alpha+1)^{a^{n}}-1 \quad \text { and } \quad a^{n+1} \|(a-1)^{\alpha^{n}}+1 .
$$

Proof. We shall prove the first relation by induction (the second is analogous). For $n=0$ the statement is obvious. Suppose that it holds for some $n$, i.e. that $(1+a)^{\alpha^{n}}=1+N a^{n+1}, a \nmid N$. Then
$$
(1+a)^{a^{n+1}}=\left(1+N a^{n+1}\right)^{a}=1+a \cdot N a^{n+1}+\binom{a}{2} N^{2} a^{2 n+2}+M a^{3 n+3}
$$
for some integer $M$. Since $\binom{a}{2}$ is divisible by a for $a$ odd, we deduce that the part of the above sum behind $1+a \cdot N a^{n+1}$ is divisible by $a^{n+3}$. Hence $(1+a)^{a^{n+1}}=1+N^{\prime} a^{n+2}$, where $a \nmid N^{\prime}$.
It follows immediately from Lemma that
$$
1991^{1993} \| 1990^{1991^{1002}}+1 \text { and } 1991^{1001} \| 1992^{1501^{1300}}-1 .
$$

Adding these two relations we obtain immediately that $k=1991$ is the desired value.
18. (BUL 1) Find the highest degree $k$ of 1991 for which $1991^{k}$ divides the number
$$
1990^{1991^{1002}}+1992^{1501^{1901}} .
$$