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 $$
n(n-1)(n-2)...(n-m+1) \cdot \frac{(n-m)(n-m-1)...1}{(n-m)(n-m-1)...1} \\
= \frac{n(n-1)(n-2)...1}{(n-m)(n-m-1)...1} \\
$$
 A: 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.
A: The notation $n!$ is used to denote the product $\prod_{j=1}^n j$ and this is commonly called "$n$ factorial." Your expression is of the form $\prod_{j=n-m + 1}^n j$ and this is thus not a factorial in this sense, as it starts at $n-m+1$ and not at $1$ (except if $n=m$). 
However, there is a pair of generalization of the notion of factorials called rising and falling factorials. 
There one has 


*

*$(n)_m$ a falling  factorial defined as $n(n-1) \dots (n-m+1)$, and 

*$(n)^m$ a rising factorial defined as $n(n+1) \dots (n+m-1)$.


The usual factorials can be recovered as the special cases $n= m$ for falling and $n=1$ for rising.
Thus, your quantity is not a factorial in the usual sense however you could conveniently express it as $(n)_m$ a falling factorial. 
For your additional question: 
The product before the fraction contains all integers from $n$ to $n+m -1$ the denominator contains all integers from $n-m$ to $1$ so taking them all together these are all the integers from $n$ to $1$, which is what is written.  
