Find all numbers whose factorial equals the product of (more than 1) consecutive numbers greater than that number. This problem is basically equivalent to the nontrivial case of the problem of finding a product of two factorials that is also a factorial, but that problem seems to be open as well. The keyword here is seems: I have no idea whether it has any solution yet. Thanks in advance for any help!
Sorry, I meant numbers. I am just extremely, extremely sorry.
 A: The factorial of the number is equal to consecutive digits, greater than the number. So $n\leq8$. We have to check those 8 cases
for $n=1$, $n=2$ there aren't solutions
for $n=3$, $3!=6 \leq 4 * 5$
for $n=4$, $4!=24 \leq 5* 6$
for $n\geq5$, the product does to have divisble with 5. The digits are greater than the number, but there aren't any digits greater than 5 divisble with 5.
So, there aren't any solutions.
If you didn't mean digits and you wanted to say numbers, we'll have to look for a different approach.
A: I have no solution but I think the problem is very interesting. So I state how far i came (note: I am not a number theorist).
At first, the number of occurences of the prime factor $p$ in $n!$ is
$$c_p(n)=\sum_{i=1}^{\infty}\left\lfloor\frac n{p^i} \right\rfloor.$$
(I can elaborate this, but I think this is not the important part.) This is roughly
$$
\sum_{i=1}^{\infty}\left\lfloor\frac n{p^i} \right\rfloor
\approx 
\sum_{i=1}^{\infty}\frac n{p^i}=\frac{n}{p-1}.
$$
For $n!\times m!=K!$ it must hold $c_p(n)+c_p(m)=c_p(K)$. By the above estimation this gives $n+m\approx K$. I guess one can do much better results by more precise estimations of $c_p(n)$.

Assume $K\geq n+m$, espcially $m\leq K-n$. Then we have
$$\frac{K!}{n!\cdot m!} \geq\frac{K!}{n!\cdot(K-n)!}={K\choose n}>1$$
for $n\not=0,K$. So we have $K<m+n$.
A: The question is to find all numbers $m$ such that there exists $j>k>1$ so that
$$m! = \frac{j!}{k!}$$
For all such cases, the numbers $j$ are listed in OEIS A109095. Every time $j$ is of the form $j=x!$ for some natural number $x\ge 3$, we have a "trivial" example:
$$(x!)! = x! \cdot (x!-1)!$$
or:
$$(x!-1)! = \frac{(x!)!}{x!}$$
For $x=3$ this gives:
$$5! = 6\cdot 5\cdot 4$$
For $x=4$ gives:
$$23! = 24\cdot 23\cdot 22\cdot\ldots\cdot 5$$
For $x=5$:
$$119! = 120\cdot 119\cdot 118\cdot\ldots\cdot 6$$
For $x=6$:
$$719! = 720\cdot 719\cdot 718\cdot\ldots\cdot 7$$
$x=7$:
$$5039! = 5040\cdot 5039\cdot 5038 \cdot\ldots\cdot 8$$
And so on.
My source for this is the OEIS entry linked above.
So the question becomes, are there other examples outside this infinite sequence?
OEIS and a comment to the question gives $10! = 6! \cdot 7!$ which holds two examples:
$$6! = 10\cdot 9\cdot 8 \text{ and } 7! = 10\cdot 9\cdot 8\cdot 7$$
So who can find other examples, or prove that no more examples exist?
The examples I mentioned are $m=5,6,7,23,119,719,\ldots$.
I found related threads, see e.g. On the factorial equations $A! B! =C!$ and $A!B!C! = D!$ and the threads link to it.
