"Double Wilson primes" - do they exist? Any prime number satisfies $(p-1)! \equiv -1 \pmod p$
Any Wilson prime number satisfies $(p-1)! \equiv -1 \pmod {p^2}$ (Only known $5, 13, 563$, widely believed that there are infinite)
Is it possible for there to exist "double Wilson primes" - those which satisfy $(p-1)! \equiv -1 \pmod {p^3}$?
And, generally, do there exist primes which satisfy $(p-1)! \equiv -1 \pmod {p^n}$?
 A: It is possible there exist such double Wilson primes, but no example is known and it is not unlikely none exists, however to prove this is likely rather hopeless. 
To see why this is the case let us recall the heuristic for the infinitude of the WIlson primes. Set $w_p = ((p-1)! +1)/p$. This is an integer, large relative to $p$.
There are $p$ possibilities for the value of $w_p$ modulo $p$ and one can assume the are all more or less "equally likely" to occur. If  $w_p=0$ modulo $p$ then we have vound a Wilson prime. 
Thus, we expect that the probability that $p$ is a Wilson prime is $1/p$. Now, $\sum_{p \le x} 1/p$  where the sum is only over the primes diverges, but slowly, namely about like $\log \log x$. 
So, we expect roughly $\log \log x$ Wilson primes up to size $x$. This is not much, and in line with the fact there are only very few known, but still the number should be infinite. 
Now, what about the double Wilson primes. Here, we need that $w_p$ is congruent to $0$ modulo $p^2$. Yet here there are $p^2$ possibilities, so we expect that the probability that $p$ is a double Wilson prime is $1/p^2$. Now, $\sum_{p } 1/p^2$ converges and thus we expect the number is finite. 
Since we have not found any and following that heuristic the expected number still to be found is really small, namely $\sum_{p \ge N} 1/p^2$ where $N$ is the value up to which one tested already, it could be there are none. But it might be some is found eventually.  
To prove anything here should be really difficult. 
