Supposing n¡ (the inverted spanish exclamation mark - as opposed to n!) uses sequential divisions, is it always true that $n! * n¡ = n^2$? Example: For n = 7, $n¡=7÷6÷5÷4÷3÷2 = 0.00972222222222222222222222222222$. If you multiply this number by 5040 (=7!) you get 49.
I've read the directions in the help center and could not understand why it is off topic. It is like asking about the relation between $x*x=x^2$ and $x÷x÷x÷x=x^{-2}$. In fact, I could not determine if this kind of question is on-topic either. And I think there is not a sister-site that would accept such kind of questions (I checked all of them). Anyways, my question has been answered. I was lazy when I failed to do some calculations to find the answer myself. This was my very first time here. I've learned something. Thank you.
Your definition amounts to $$ n¡ = \frac{n}{(n-1)!} $$ so for instance, we would compute $7¡ = 7/(6!)$. Clearly, then, we have $$ (n!) \cdot (n¡) = [n \cdot (n-1)!] \cdot \frac{n}{(n-1)!} = n^2 $$