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Solve $(n+1)^k-1=n!$ in positive integers.

So far, I can see that $n! = -1$ (mod $n+1$), which is true for all $n+1$ prime. Also, $(1,1), (2,1), (4,2)$ are solutions. They seem like special cases, and I might need to prove something about mod $5$, since I do not see solutions above that. However, I cannot continue.

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  • $\begingroup$ There are three small solutions $\endgroup$
    – Henry
    Nov 17, 2021 at 12:34
  • $\begingroup$ Thanks, editing the post $\endgroup$
    – xrider1000
    Nov 17, 2021 at 12:38
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    $\begingroup$ $n+1$ must be prime. $\endgroup$
    – Barry Cipra
    Nov 17, 2021 at 12:42
  • $\begingroup$ May I know why is that so? $\endgroup$
    – xrider1000
    Nov 17, 2021 at 12:43
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    $\begingroup$ $n+1$ must be prime since $n! \equiv -1 \pmod{n + 1}$. See Wilson's theorem. $\endgroup$
    – VTand
    Nov 17, 2021 at 12:45

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