I'm pretty sure the following statement is true:
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 $n$.
But I'm having trouble proving it.
Here is an attempt by induction. The property is true when $n=6$, and assume it's true for $n$. If $n+1$ isn't prime, the induction step is trivial, for the smallest prime bigger than $n+1$ is equal to the smallest prime bigger than $n$; call this prime $p$. But by hypothesis all composites smaller than $2p$ divide $n!$, hence $(n+1)!$.
The harder case is when $n+1$ is prime. Let $q$ denote the next prime, i.e. the smallest prime bigger than $n+1$. We know by hypothesis that all composites smaller than $2(n+1)$ divide $n!$, hence $(n+1)!$. We also know $2(n+1)$ divides $(n+1)!$. To finish, we need to show that all composites $m$ strictly between $2(n+1)$ and $2q$ divide $(n+1)!$.
This is where I get stuck. It certainly helps that the ratio $\frac{q}{n+1}$ can't be larger than 2 (by Bertrand's postulate; I imagine the bound can be sharpened but I know embarrassingly little number theory). It's also obvious that the prime factors of any such composite $m$ are all smaller than $q$. What I don't quite see is an argument to ensure the powers of those prime factors aren't too large.
Feel free to give alternative approaches, rather than by induction, if there is a much simpler proof I've overlooked.