1. Find all pairs of positive integers $(n . k)$ so that $(n+1)^{k}-1=n$ !.
1. If $p$ is a prime factor of $n+1$, it is also a factor of $n!+1$. However, none of the numbers $1, \ldots, n$ is a factor of $n!+1$. So we must have $p=n+1$, i.e., $n+1$ must be prime. For $n+1=2,3,5$, we have solutions $(n, k)=(1,1),(2,1)$ and $(4,2)$. Let us show that there are no other solutions. Indeed, suppose that $n+1$ is a prime number $\geq 7$. Then $n=2 m, m>2$. Now $2n^{k} \geq n^{n} \geq n!,
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a contradiction.