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My main question is similar to the title:

Are there any primes $p$ such that $p^3$ divides $(p-1)!+1$?

It is hard to find all $p$ such that $p^2$ divides $(p-1)!+1$ (Wilson primes).

So, in my opinion, there might be no primes $p$ such that $p^3$ divides $(p-1)!+1$. How can I prove this? If it is incorrect, how can I find a prime that satisfies the condition?

Let me know if this question is appropriate or not. If it is inappropriate, I will delete it immediately.

Edit: Thank you @FrançoisBrunault for the link to a similar question: Stronger versions of Wilson's Theorem

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    $\begingroup$ I suspect the answer to all three of your questions is, nobody knows. $\endgroup$
    – Gerry Myerson
    Commented Sep 10, 2019 at 7:26
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    $\begingroup$ The usual heuristic is that the "probability" that a prime $p$ satisfies $p^3 \mid \big( (p-1)!+1\big)$ is $1/p^2$. Since there are no such primes up to $2\times10^{13}$ (as per the Wikipedia article on Wilson primes), our heuristic is that there is a (much) less than $1$ in $2\times10^{13}$ chance that any such primes exist. But yes, a proof seems hopeless at present. $\endgroup$
    – Greg Martin
    Commented Sep 10, 2019 at 7:53

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