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.
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.
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 $K<m+n$.
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.
