Proposition: Let $W(1)$ be the set of all Wilson primes of order $1$ and suppose $n=p-1,$ where $p$ is a prime such that $p\notin W(1)$, then there are no integer solutions to the equation $$n!+1=m^2$$.
Proof:
Wilson's theorem: $(p-1)!\equiv-1 \pmod p \tag 1$
$$$$ Using $(1)$, we know that $(p-1)!+1=ap$ for some $a\in\mathbb{N}$.
In order for the product $ap$ to be a perfect square, $a$ must be of the form $k^2p$ for some $k\in\mathbb{N}$.
So, $(p-1)!+1=k^2p^2$
But for primes $p\notin W(1)$, $(p-1)!\not\equiv-1 \pmod {p^2}$.
Therefore, $(p-1)!+1$ cannot be represented as a square of an integer. $\square$
$$$$The above is a glaringly obvious implication (although very weak) of Wilson's theorem on Brocard's Problem. However, I cannot find any reference to it in literature. Am I missing something?
