3. Find all pairs of positive integers $(n, k)$ such that $(n+1)^{k} -1=n!$.

 untranslated portion:
-1=n!. 

(Note: The last part "-1=n!." was already in English and thus not translated.)
3. Solution: That is, $(n+1)^{k}=n!+1$.

If $p$ is a prime factor of $n+1$, then $p$ is also a prime factor of $n!+1$,
$\because$ any prime $q \backslash n!+1(q \leqslant n)$,
$\therefore p=n+1$, if $n+1=2$, then $n=1, k=1$;
if $n+1=3$, then $n=2, k=1$; if $n+1=5$, then $n=4, k=2$.

If $(n, k)$ is a solution, then $n+1$ is a prime number greater than 7, we can set $n=2 m(m>2)$,
$\because 2n^{k} \geqslant n^{n}>n!$, contradiction. Therefore, all solutions are $(1,1),(2,1),(4,2)$.