For $n\geq 6$, the smallest composite number that is not a factor of $n!$ is $2p$, where $p$ is the smallest prime bigger than $n$?

A composite $m$ with $2(n+1)\lt m \lt 2q$ cannot be twice a prime because $q$ is the next prime after $n+1$, so must be able to be factored into $ab$ with $a,b \lt n+1$. Then $a,b$ are separately factors of $(n+1)!$ and $ab$ divides $(n+1)!$ unless $a=b$. We will have $a^2$ divide $(n+1)!$ unless $a \gt \frac {n+1}2$ because $a,2a$ are both factors. But then $a^2 \gt \frac {(n+1)^2}4 \ge 4(n+1) \ge 2q$ as long as $n \ge 15$ by Bertrand's postulate. We can check the cases up to $15$ by hand to complete the proof.

In my opinion, induction is not always the best way to prove everything.

For this problem, I would first review the basic definition of the factorial: $$n! = \prod_{i = 1}^n i.$$ This means that $n!$ is divisible by every prime less than $n$, e.g., $6!$ is divisible by $2, 3, 5$ but not by $7$. Now, $8 \mid 6!$, as also do $9, 10, 12$.

That's because they're composite numbers divisible by some combination of the primes that divide $6!$ but without excess multiplicity, e.g., $8 = 2 \times 4$, $9 \mid 3 \times 6$, $10 = 2 \times 5$, $12 = 3 \times 4$.

Obviously $p \nmid n!$ if $p$ is the smallest prime greater than $n$. But $p$ is prime, not composite. If there are composite numbers between $n$ and $p$, it seems obvious to me that they must be divisible by some prime less than $p$.