He says in proving Wilson
"If $n$ is not a prime, then its factors are to be found among the set of integers $\{1, 2, . . . , n-1 \}$, and so $(n-1)!$ is divisible by $n$, that is, $(n-1)!= 0 (\text{mod} \ n)$"
but for $n = 4 = 2 \cdot 2$ we have $(4 - 1)! = 3! = 6$ which is not divisible by $n = 4$ unless I'm misunderstanding basic number theory (again) :p