Factorials and prime numbers

Proof of (C): let $N<m<N!$ be the smallest divisor of $(N!-1)$. We argue by contradiction: if $m$ is not prime, then it has some nontrivial divisor $1<m'<m$, and thus $m'$ also divides $(N!-1)$. By minimality of $m$ it follows that $m'\le N$, but no such number can divide $(N!-1)$.