$n! =$ the product of consecutive integers. Can $n!$ be the product of $k$ consecutive integers for $k > 1$? (Not including the degenerate cases such as when $k = 2$, then $1\cdot2 = 2!$ and $2\cdot 3 = 3!$, and so on.)
I am asking not for $n!$ to be divisible by $n$ consecutive integers, I am asking for $n!$ to equal the product of $k$ consecutive integers, implying that $n$ is not necessarily equal to $k$ (And when $n = k$, then clearly there exists an answer, namely $n! = (1)(2)(3)...(k)$)

The formula $$(n!-1)!=\frac{(n!)!}{n!}=\prod_{k=n+1}^{n!} k$$ gives an infinite family of examples.
