2. (Finnish High School Mathematics Competition 2002-2003) Find all positive integer pairs $(n, k)$ that satisfy $(n+1)^{k}-1=n$!. 

 untranslated text:
 求满足 $(n+1)^{k}-1=n$ ! 的所有正整数对 $(n, k)$. 

 translated text:
 Find all positive integer pairs $(n, k)$ that satisfy $(n+1)^{k}-1=n$!.
2. If $p$ is a prime factor of $n+1$, then $p$ is also a prime factor of $n!+1$. Since $n!+1$ cannot be divisible by any prime $q(q \leqslant n)$, it must be that $p=n+1$.
If $n+1=2, n=1$, and $2^{k}-1=1$, then $k=1$, so the pair $(1,1)$ is a solution.
If $n+1=3$, similarly $3^{k}-1=2 \Rightarrow k=1$. Therefore, the pair $(2,1)$ is a solution.
If $n+1=5$, similarly $5^{k}-1=24 \Rightarrow k=2$. Therefore, the pair $(4,2)$ is a solution.
Below we prove there are no other solutions.
Assume $(n, k)$ is a solution satisfying the condition, and $n+1$ is an odd prime greater than or equal to 7, then:
$$
n=2 m, m>2 \text {. }
$$

Since $2 \leqslant n-1, m \leqslant n-1, n=2 m$ is a factor of $(n-1)!$, so $n!$ can be divided by $n^{2}$. Therefore, $(n+1)^{k}-1=$ $n^{k}+k n^{k-1}+\cdots+\frac{k(k-1)}{2} n^{2}+n k$ can be divided by $n^{2}$, thus $k$ can be divided by $n$, hence $k \geqslant n, n!=(n+1)^{k}-1>$ $n^{k} \geqslant n^{n}>n$ ! contradiction.