Infinitely many primes that $\exists n, p\mid n!+1, n\nmid p-1$ Prove that there are infinitely many primes $p$ such that, there exists a positive integer $n$ such that $p\mid n!+1$ and $n \nmid p-1$.
This seems to be a generalization of Wilson's theorem. From first glance, $n$ must be less than $p-1$ as otherwise $p\mid 1$. Assuming only finitely many such primes doesn't seem to give any useful information that relates to Wilson's theorem.
 A: I also don't see how to use an assumption of only finitely many such primes. Instead, consider
$$k = 6j + 1, \; \; j \in \mathbb{N}^{+} \tag{1}\label{eq1A}$$
For any particular $j$, and thus $k$, if there's at least one prime $p$ with $p \mid k! + 1$ and $k \nmid p - 1$, we can use that prime $p$, with $n = k$. Otherwise, then all prime factors $p$ of $k! + 1$ have $k \mid p - 1$. Choosing any one of these primes, since $2k + 1 = 2(6j+1) + 1 = 12j + 3$ is divisible by $3$, there is then an even integer $m$ where
$$p = mk + 1, \; \; m \ge 4 \tag{2}\label{eq2A}$$
Next, using $k$ being odd and $k! \equiv -1 \pmod{p}$, we have
$$\begin{equation}\begin{aligned}
\prod_{i=p-k}^{p-1}i & \equiv \prod_{i=p-k}^{p-1}(-1)(p-i) \\
& \equiv (-1)^{k}\prod_{i=p-k}^{p-1}(p-i) \\
& \equiv (-1)\prod_{i=1}^{k}i \\
& \equiv (-1)k! \\
& \equiv 1 \pmod{p}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Going from the second to the third line's product, the order of factors was reversed and the indices were adjusted. Using \eqref{eq3A} and Wilson's theorem, we then get
$$\begin{equation}\begin{aligned}
\prod_{i=1}^{p-1}i & \equiv -1 \pmod{p} \\
\left(\prod_{i=1}^{p-k-1}i\right)\left(\prod_{i=p-k}^{p-1}i\right) & \equiv -1 \pmod{p} \\
\prod_{i=1}^{p-k-1}i & \equiv -1 \pmod{p} \\
(p-k-1)! & \equiv -1 \pmod{p}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Thus, using \eqref{eq2A}, $p \mid ((m-1)k)! + 1$. However, if $(m-1)k \mid mk$, then $m - 1 \mid m$, but this isn't possible since $\gcd(m-1,m) = 1$ and $m-1 \gt 1$. Thus, we can use this prime $p$ with $n = (m-1)k$.
As you mentioned, all primes $p$ where $p \mid n! + 1$ have $p \ge n+1$. Therefore, with $k$ increasing without bound with increasing $j$, the above shows there are infinitely many primes $p$ that meet the conditions.
