1990 AIME Problem 11

Compare $n!$ with products of $n-3$ consecutive integers beginning at $4$ and at $5$

The product beginning at $5$ equals $\frac{(n+1)!}{4!}$

The $n-3$ consecutive integers beginning at $4$ have product $$4\cdot5\cdots n=\frac{n!}{6},$$ while those beginning at $5$ have product $$\begin{aligned}5\cdot6\cdots(n+1)&=\frac{(n+1)!}{24}\\&=\frac{n+1}{24}n!.\end{aligned}$$ For $n=23,$ the latter product equals $n!,$ so $23$ works. For every $n\geq24,$ the product beginning at $4$ is below $n!,$ the product beginning at $5$ is above $n!,$ and the product strictly increases with its initial term. Hence no $n\geq24$ works, and the largest possible value is $23.$