# Factorials and anti-factorials 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.

Factorials and anti-factorials
Supposing n¡ (the inverted spanish exclamation mark - as opposed to n!) uses sequential divisions, is it always true that $n!\ast n¡={n}^{2}$? 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 $x\ast x={x}^{2}$ 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.
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empatiji2v
$\frac{n}{\left(n-1\right)!}\cdot n!={n}^{2}$

Leroy Gray
$n¡=\frac{n}{\left(n-1\right)!}$
so for instance, we would compute $7¡=7/\left(6!\right)$. Clearly, then, we have
$\left(n!\right)\cdot \left(n¡\right)=\left[n\cdot \left(n-1\right)!\right]\cdot \frac{n}{\left(n-1\right)!}={n}^{2}$