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

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

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For a more accurate and complete translation, the last part should be:

Find all positive integer pairs $(n, k)$ that satisfy $(n+1)^{k}-1=n$!.
10. 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 follows that $p=n+1$.
If $n+1=2, n=1$ and $2^{k}-1=1$, then $k=1$. Therefore, the pair $(1,1)$ is a solution.
If $n+1=3$, similarly, we get $3^{k}-1=2$ and $k=1$. Therefore, the pair $(2,1)$ is a solution.
If $n+1=5$, similarly, we get $5^{k}-1=24$ and $k=2$. Therefore, the pair $(4,2)$ is a solution.
Below we prove that 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 $n=2 m, m>2$.
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 $n$, and $k \geqslant n$, hence, $n!=(n+1)^{k}-1>n^{k} \geqslant n^{n}>n!$. Contradiction.