TITLE: $n!$ as product of consecutive numbers
QUESTION [14 upvotes]: 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.

REPLY [3 votes]: There are infinitely many $n$ for which $n!$ can be written in 3 ways. All you have to do is set $n_k = 2 \cdot 3 \cdot 4 \ldots \cdot k$. Then whenever $n = n_k - 1$ for some $k$, you can remove the first $k$ numbers in the product and add $n_k$ to the end. It seems much more complicated to prove that you can add, say, two extra consecutive numbers to the end of the product for infinitely many $n$ (and determine for which $n$ when you can, in analytic or fast computational form). If you add two numbers to the end, $n+1$ and $n+2$, then $GCD(n+1,n+2) = 1$ so somehow $n+1$ and $n+2$ must partition the prime factors of $n_k!$. I'm currently running a computer search to find the first $n$ (if any) where this is possible. I'll update if I find anything.