We have that $1\cdot2\cdot3\cdot4\cdot5\cdot6=8\cdot 9\cdot10$. An easy consequence is that $7!=7\cdot8\cdot9\cdot10$.
I have been looking for more non trivial examples like these, but I have found none. Is there some known fact respect to this?
More specifically: has the equation $$\frac{j!}{k!}=\frac{m!}{n!}$$ more non trivial solutions, other than $(j,k,m,n)=(6,1,10,7)$ and $(j,k,m,n)=(7,1,10,6)$?
What do you mean by trivial? Here are some more pairs:
$$ \frac{n!}{1!} = \frac{(n!)!}{(n! - 1)!} $$
Simple python code gives little bit more interesting pairs:
$( 6, 3, 5, 1 ), ( 10, 6, 7, 1 ), ( 12, 4, 11, 2 ), ( 15, 7, 13, 4 ), ( 20, 5, 19, 3 ), ( 24, 4, 23, 1 ), ( 30, 6, 29, 4 ) , ( 42, 7, 41, 5 ), ( 56, 8, 55, 6 ), ( 57, 22, 54, 18 ), ( 60, 5, 59, 2 ) , ( 66, 14, 62, 7 ), ( 72, 9, 71, 7 )$.
Here's my write-up of the general problem:
The "Puzzle Corner" of MIT News for March/April 2020 gives a "speed" problem by Sorab Vatcha: "Find seven consecutive integers whose product equals the product of four consecutive integers." The obvious "speed" solution is: 0, 1, 2, 3, 4, 5, and 6; and 0, 1, 2, and 3.
This leads to the general question of whether there are any "nontrivial" solutions to find two distinct sequences of integers whose products are the same. The trivial solutions involve (1) sequences containing 0, (2) replacing all of the integers in a sequence with their negatives, (3) adding or deleting values 1 or -1. This reduces the problem to find two distinct sequences of integers $\ge 2$ of length $\ge 2$ whose products are the same.
There are solutions to this. The smallest product is probably $2\ 3\ 4\ 5 = 4\ 5\ 6 = 120$. Is there an enumeration of all solutions?
It appears that there are an infinite number of solutions, but nobody has presented an infinite family of nontrivial solutions.
Edit to add:
We can add a further criterion of non-triviality: The two sequences must not overlap. If the sequences overlap, then the overlapping part can be deleted from both of them, yielding a soltion with shorter sequences. This interacts with the prohibition against sequences of length 1: Removing the overlapped part may reduce one sequence to length 1, and the shorter solution may be trivial also. And indeed, there is a large family of solutions constructed this way: If the product of $a \cdots b$ is $P$, then $a \cdots (P-1) = (b+1) \cdots P$. (The solution I gave above has this form.)
I've run a computer search for products up to $10^{20}$, and the only solutions are $5 \cdots 7 = 14 \cdot 15 = 210$, $2 \cdots 6 = 8 \cdots 10 = 720$, $19 \cdots 22 = 55 \cdots 57 = 175560$, and $8 \cdots 14 = 63 \cdots 66 = 17297280$.
Yes, there are other solutions, such as
$$\frac{3!}{1!}=\frac{6!}{5!}$$
$$\frac{11!}{2!}=\frac{12!}{4!}$$
$$\frac{13!}{4!}=\frac{15!}{7!}$$
As shown in the other answer, the equation in fact has infinitely many solutions, though as the last two examples above illustrate, not all of them are generated by the formula given in that answer.