Isn't $ n(n-1)(n-2)...(n-m+1) $ a factorial already? Let $ m \ge 1 $ and $ n \ge 1 $ be integers
Let $A$ be a set of size $m$
Let $B$ be a set of size $n$
How many one-to-one functions $f: A \rightarrow B$ are there?
skipped stuff
$$
n(n-1)(n-2)...(n-m+1) \\
= n(n-1)(n-2)...(n-m+1) \cdot \frac{(n-m)(n-m-1)...1}{(n-m)(n-m-1)...1} \\
= \frac{n(n-1)(n-2)...1}{(n-m)(n-m-1)...1} \\
= \frac{n!}{(n-m)!}
$$
My question is: Isn't $ n(n-1)(n-2)...(n-m+1) $ already a factorial? Also isn't the $+1$ there to make it so you don't multiply against $0$ for product rule?
Also how does $$
n(n-1)(n-2)...(n-m+1) \cdot \frac{(n-m)(n-m-1)...1}{(n-m)(n-m-1)...1} \\
= \frac{n(n-1)(n-2)...1}{(n-m)(n-m-1)...1} \\
$$

No, a factorial is something like $1\cdot2\cdot3\cdot4\cdot5$. You have something like $9\cdot8\cdot7\cdot6\cdot5$. That is, a factorial has to start from $1$ and go up - yours starts from some number and goes down (and crucially, doesn't go down all the way to $1$, otherwise it would be a factorial, just written backwards).
However, you can express products like that using factorials, which is what they're doing here. For example:
$$9\cdot8\cdot7\cdot6\cdot5=9\cdot8\cdot7\cdot6\cdot5\cdot\not4\cdot\not3\cdot\not2\cdot\not1=\frac {9!} {4!}$$
The $+1$ is not really there "to avoid multiplication by $0$", it's there because... well, because the formula is incorrect without it.
