jmc

The product of $n-3$ consecutive integers can be written as $\frac{(n-3+a)!}{a!}$ for some integer $a$. Thus, $n!=\frac{(n-3+a)!}{a!}$, from which it becomes evident that $a\ge 3$. Since $(n-3+a)!>n!$, we can rewrite this as $\frac{n!(n+1)(n+2)\dots (n-3+a)}{a!}=n!⟹(n+1)(n+2)\dots (n-3+a)=a!$. For $a=4$, we get $n+1=4!$ so $n=23$. For greater values of $a$, we need to find the product of $a-3$ consecutive integers that equals $a!$. $n$ can be approximated as ${}^{a-3}\sqrt{a!}$, which decreases as $a$ increases. Thus, $n=\boxed{23}$ is the greatest possible value to satisfy the given conditions.