Question: I want to derive why this equation is true.

$$\frac{q}{r} \times p(q, r) = P(q-1, r)- P(q-1, r+1) $$

, where

$$p(q, r) = \frac{r(r+1)(r+2)\dots(r+q-1)}{q!}p^q(1-p)^r$$,

$$P(q-1, r) = \sum\_{x=0}^{q-1} \frac{r(r+1)(r+2)\dots(r+x-1)}{x!}p^x(1-p)^{r}$$

$$P(q-1, r+1) = \sum\_{x=0}^{q-1} \frac{(r+1)(r+2)\dots(r+x)}{x!}p^x(1-p)^{r+1}$$

I've been looking at it for a while, but I don't seem to get it.

I started off writing the equation as:

$$\frac{(r+1)(r+2)\dots(r+q-1)}{(q-1)!}p^q(1-p)^r = \sum\_{x=0}^{q-1} \frac{r(r+1)(r+2)\dots(r+x-1)}{x!}p^x(1-p)^{r}-\sum\_{x=0}^{q-1} \frac{(r+1)(r+2)\dots(r+x)}{x!}p^x(1-p)^{r+1}$$

But I am not getting any further besides endlessly manipulating things, but that doesn't brought success.

Any hints are welcome!

Answer: 
If you know that

1. $-a=(-1)\cdot a$
2. $-(-a)=a$

Note that

$1=\_2-(-1)=\_1(-1)\cdot (-1)$
You can also see that $x^2=1$ has two solutions $1$ and $-1$ since a field is necessarily an integral domain.