These are the numbers that can be written in three different ways less than $700,000$:
$120=4\cdot5\cdot6=2\cdot3\cdot4\cdot5=1\cdot2\cdot3\cdot4\cdot5$
$720=8\cdot9\cdot10=2\cdot3\cdot4\cdot5\cdot6=1\cdot2\cdot3\cdot4\cdot5\cdot6$
$5040=7\cdot8\cdot9\cdot10=2\cdot3\cdot4\cdot5\cdot6\cdot7=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7$
Additionally, any number of the form $(n!-1)!$ can be expressed in 3 ways. Specifically, $(n!-1)!$, $(n!-1)!/1$, and $n!(n!-1)!/n!$. An example will help explain.
For $n=15$, we have $1\cdot2\cdots(15!-1)=2\cdot3\cdots(15!-1)=16\cdot17\cdots(15!-1)\cdot(15!)$
These are very, very large numbers, where $n=4$ results in $2.58\cdot10^{22}$, and $n=5$ results in $5.57\cdot10^{196}$, but there are infinitely many.
I don't believe I can do the same trick to get a product of consecutive integers in $4$ ways since $n!+1$ doesn't have anything to do with $(n+1)(n+2)\cdots(n+k)$. Maybe some kind of triple factorial but that would have the same issue I fear.
I've learned about the proof regarding when $a(a+1)=b(b+1)(b+2)$ has integer solutions, and how it has a finite number of solutions, but what I don't know is how I would generalize this since this is a much worse form Diophantine equation:
$$x=\prod_{n=0}^{k_1-1}(a_1+n)=\prod_{n=0}^{k_2-1}(a_2+n)=\prod_{n=0}^{k_3-1}(a_3+n)=\prod_{n=0}^{k_4-1}(a_4+n)$$
For distinct integers $k_1<k_2<k_3<k_4$.
We can see clearly that $x$ must be divisible by $60$ since $\{k_1,k_2,k_3,k_4\}=\{2,4,5,6\}$ is the simplest solution possible and if $k_4>6$, it will still be divisible by lcm$(2,3,\cdots,k_4)$, which is divisible by $60$.
$x\ge k_4!>$ lcm$(2,3,\cdots,k_4)$
I think another approach may be to look at the prime factors of the number and see that if $p$ is the largest prime that divide $x$, then any product will have to go through $p$ since $2p$ would be unlikely to be in a sequence of composite numbers that multiply to $x$. Not sure how to express this mathematically.
As a notation question, is it okay to use the same variable $n$ in the $\prod$'s above? I figured it was a little easier to read.
Edit: this is related to A100934 and A064224 in OEIS. It appears that it is an open question that if $n$ is not a factorial number if it can be expressed in more than $2$ different ways. Note the trick where you remove the $1$ is only possible with factorial numbers.
In regards to both factorial and non-factorial: "MacLeod and Barrodale prove that the equation $x(x+1)\cdots(x+m-1) = y(y+1)\cdots(y+n-1)$ has no solutions $x>1$ and $y>1$ for the following pairs of $(m,n): (2,4), (2,6), (2,8), (2,12), (4,8), (5,10)$. They also show that $(2,3)$ has two solutions and $(3,6)$ has one solution. They conjecture that $(2,k)$ has no solution for $k>3$. [T. D. Noe, Jul 29 2009]"
In reference to non-factorial elements in A064224: "The early terms in this sequence each have two representations. Is two the maximum possible? The sequence is infinite: for any n, the number $n\cdot(n+1)\cdots(n^2+n-1)$ is in this sequence. The next number of this form is $20,274,183,401,472,000$, which is obtained when $n=4$. - T. D. Noe, Nov 22 2004"
