So I am working on factorials and can easily show that $$r = \frac{r!}{(r-1)!}$$
By expanding the factorial $$r! = r(r-1)\cdots(r-n+1)$$ Then dividing both sides by $(r-1) \cdots (r-n+1)=(r-1)!$
However, I cannot figure out the equivalent for $$r-n+1 = \frac{r!}{r(r-1) \cdots(r-n)}$$
Oh, I get what you are asking!
You are asking what does $r(r-1)\cdots(r-n+1)$ equal in a simple factorial equation.
Well. $r!$ is $r$ multiplied by everything down to $1$. But $r(r-1)\cdots(r-n+1)$ is $r$ multiplied only down to $r-n+1$. Everything else $r-n$ down to $1$ is missing. If we multiplied by $(r-n).......2\cdot 1$ we'd get.
$r! = [r(r-1)\cdots(r-n+1)]\times [(r-n)...... 2\cdot 1]$.
Note: $[(r-n)...... 2\cdot 1] = (r-n)!$ so if we divide both sides by that we get:
$\frac {r!}{(r-n)!} = [r(r-1)\cdots(r-n+1)]$
This is useful as shorthand to think whenever you see $\frac {r!}{(r-n)!}$ that that means $r(r-1).........(r-n+1)$.
This way if anyone ask you "If you have $52$ cards and you deal out $17$ of them then how many different sequence of $17$ cards can you have" you can say "Well there are $52$ choices for the first dealt card and $51$ choices for the second all the way down to $36$ choices for the $17$th card so that is $52\cdot 51\cdot .... \cdot 36 = \frac {52!}{35!}$". Now you can know you can just take a shortcut... There are $\frac {52!}{(52-17)!}$ ways to do it.