Factorials and prime numbers

Question

Let $N$ be a positive integer not equal to $1$. Then note that none of the numbers $2, 3, \ldots, N$ is a divisor of $N! - 1$. From this we can conclude that:

(A) $N! – 1$ is a prime number;

(B) at least one of the numbers $N + 1, N + 2, \ldots, N! – 2$ is a divisor of $N! - 1$;

(C) the smallest number between $N$ and $N!$ which is a divisor of $N! - 1$, is a prime number.

My working: $N! - 1$ is not necessarily a prime, as $5! - 1 = 119$ and $7 \mid 119$. I cannot proceed further. Hints are appreciated.

Answer 1

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 $(N!-1)$.

Answer 2

(A) is a premature conclusion. If $N! - 1$ is composite, its least prime factor could be as large as $\sqrt{N! - 1}$. And if $N > 4$, then $\sqrt{N! - 1} > N$, as in, for example, $\sqrt{119} \approx 10.9 > 5$.

(B) contradicts (A), but it seems like it could be a more useful conclusion if we adjust it to exclude the "factorial primes". However, since $N! - 2 > \sqrt{N! - 1}$ by a rapidly widening margin, it turns out to be overkill, e.g., $8! - 1 = 40319 = 23 \times 1753$, and clearly $\sqrt{40319} < 40318$.