Q: If $n\equiv (p-1)!-1 \ \mbox{mod p!}$ then the $p-2$ integers that precedes $n$ and the $p$ integers that succeeds $n$ are composite. 
Let $p$ prime, $p>3.$ If $n\in \mathbb N\setminus \{0\}$ is such that $n\equiv (p-1)!-1 \ (\mbox{mod p!})$, then the $p-2$ integers that precedes $n$ and the $p$ integers that succeeds $n$ are composite.

I've proven that $\mbox{gcd}((p-1)!-1,p!) =1$. In fact, given any common divisor $c$, we have $c|p! $ and $c|(p-1)!-1$. Therefore, $c| p! - p(-1+(p-1)!)= p!+p-p! = p.$ Since $p$ is prime, it must be $c=1$.
Ok, now by hypothesis we have $p!|n-((p-1)!-1)$. Since we've proven that $p!\not| (p-1)!-1,$ it follows that $p!\not|n$. 
What now? How do i proceed with those results? Thanks in advance!

A: I am not sure you are going in the right direction. However, here is what you can do:
[1] The problem should be fairly easy for $n+k$ where $k\in\{-(p-2),...,-1,3,...,p\}$, because $n+k \equiv (p-1)! + k-1 (\mod p!)$, with $k-1 \in \{-(p-1),...,-2,2,...,p-1\}$ so $k-1|n+k$, and it is a proper divisor because of the size of it. ($n+k \ge (p-1)!-(p-1) \gt p $ for $p\gt 3$.) What is left are the cases $k=1$ and $k=2$.
[2] If $k=1$, then $n+k \equiv (p-1)! (\mod p!)$ so $n+k$ is divisible by all numbers from 2 up to $p-1$, and those are proper factors as in the previous case.
[3] If $k=2$, then $n+k \equiv (p-1)!+1 (\mod p!)$. It is a well-known result (Wilson's theorem, https://en.m.wikipedia.org/wiki/Wilson%27s_theorem) that, for prime $p$, $(p-1)! \equiv -1 (\mod p) $. This and the previous equivalence mean that $n+k$ is divisible by $p$, and again, by size argument, you conclude that $p$ is its proper factor.
