I am attempting to place the following finite product of terms into a binomial, for example: $$A'=\frac{A(A+1) \cdots (A+n-2)}{(s+A-1)(s+A-2) \cdots (s+A-n+1)}$$
I recognise the product on the numerator goes to $n-2$, which may look like $$\prod_{i=0}^{n-2}(A+i)$$ and the product on the numerator becomes the following $$\prod_{i=1}^{n-1}(s+A-i)$$
And combining the two I then have $$A' = \frac{\prod_{i=0}^{n-2}(A+i)}{\prod_{i=1}^{n-1}(s+A-i)}$$
For the binomial form, I get the following $$A' = \frac{\binom{A+n-2}{n-2}}{\binom{s+A-n+1}{-n+1}}$$
The text from which I got this from, shows the binomial form to be $$\frac{\binom{n+A-2}{n-1}}{\binom{s+A-1}{n-1}}$$
The denominator does not look correct to me
The numerator of $A'$ is just $$A(A+1)\cdot \ldots \cdot (A+n-2) = \frac{(A+n-2)!}{(A-1)!}.$$ Similarly, the denominator of $A'$ is $$(s+A-1)(s+A-2)\cdot \ldots \cdot (s+A-n+1) = \frac{(s+A-1)!}{(s+A-n)!}.$$
So we must have $$A' = \frac{(A+n-2)!}{(A-1)!} \frac{(s+A-n)!}{(s+A-1)!}.$$
Now for a binomial coefficient of the form $$\binom{n}{k} = \frac{n!}{k!(n-k)!},$$ we must have the denominator arguments adding up to the numerator; i.e., $k + (n-k) = n$. So one way to achieve this is to write
$$\frac{(A+n-2)!}{(A-1)!} = \frac{(A+n-2)!}{(A-1)!(n-1)!} (n-1)! = \binom{A+n-2}{n-1} (n-1)!, \\ \frac{(s+A-n)!}{(s+A-1)!} = \frac{(s+A-n)!(n-1)!}{(s+A-1)!} \frac{1}{(n-1)!} = \frac{1}{\binom{s+A-1}{n-1}} \frac{1}{(n-1)!}.$$
Therefore $$A' = \frac{\binom{A+n-2}{n-1}}{\binom{s+A-1}{n-1}}.$$ Hence