One way to show that there are infinitely many primes is to show that for $\forall n \ge 3$, there is a prime p such that $n \lt p \le n!-1.$
I thought of assuming that the numbers in the sequence $n+1, n+2,... n!-1$ are all composite and deducing a contradiction, but I don't know how to do this.
Let $p_1, p_2, \dots p_k$ be all the primes in $\{1, 2, \dots, n\}$, and let $P=p_1p_2\dots p_k$. Of course $P\le n!$, so $P-1\le n!-1$ and $P-1$ is not divisible by $p_1, p_2, \dots, p_k$, therefore there must be a prime $p>n$ that divides $P-1$, so $p$ is a prime that satisfies $n<p\le P-1\le n!-1$.
Note that the statement is true for $n = 3$ and $n = 4$. In particular, $5$ and $7$ are prime and both $3 < 5 \leq 3! - 1 = 5$ and $4 < 7 \leq 4! - 1 = 23$.
Now suppose $n$ is a positive integer with $n \geq 5$ and that none of $n+1$, $n+2$, ... $n!-1$ is prime. Set $Q$ to be the product of the primes in $[1,n]$. Since $4 < n$, $Q < n!$ and $Q - 1 < n!-1$. We know $Q-1$ can be factored into a product of primes. Let $q$ be any prime in $[2,n]$. $q$ does not divide $Q$ because $Q/q$ has remainder $q-1 \neq 0$. So all the prime divisors of $Q$ are in $[n+1, n!-1)$. Therefore, there is at least one prime, $p$, dividing $Q-1$ with $n < p \leq n!-1$.
