$n!$ 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.
 A: 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 
$$\frac{(m!)!}{m!}=(m!-1)!$$
For instance
$$5\cdot6\cdots24=1\cdot2\cdots23.$$
and there are infinitely many solutions, extremely large.
Other solutions are accidental and rare.
A: 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.
A: It exists at least 2 trivial combinations (1*2*...*n and 2*3*...*n). 
It will also exist a third combination if n+1 is the potencial of other number, although it is not an enough condition.
For example:
while 23! = 1*2*3*...*23 = 2*3*4*5*...*23 = 5*6*...*23*24
To get a fourth combination continuing this method, it would have to be a number that reach next conditions:


*

*n+1 = m!

*n+2 = p!/m!


but as soon as p!/m! is always an even number (p>m+1), the difference between it and m! (also even) cannot be one.
For that reason, it is not possible to have 4 combinations.
