# n!n! as product of consecutive numbers

Let $n$ be a positive integer.

In how many ways can one write $n!$ as a product of consecutive integers?

For example: $4!=1\times2\times3\times4=2\times3\times4$. Here, $2$ possibilities exist.

$5!=1\times2\times3\times4\times5=2\times3\times4\times5=4\times5\times6$. Here, $3$ possibilities exist.

The products of consecutive integers, $(m+1)\cdot (m+2)\cdot(m+3)\cdots n$ is a factorial from which you have removed the first factors, i.e $\dfrac{n!}{m!}$. To make this equal another factorial, you need to discard the largest factors of $n!$, i.e. $n\cdot(n-1)\cdot(n-2)\cdots$.
It is always possible to discard $n$ itself. This yields