Pascal's Identity: ${n \choose {r-1}} + {n \choose r} = {{n+1} \choose r}$?

Here's the proof:

(1)  The definition of the binomial coefficient is:

$${n \choose r} = \frac{n(n-1)\times\dots\times(n-r+1)}{r!}$$

(2)  This analysis will assume that $n,r$ are integers.

(3)  If $r > n+1$, it follows that Pascal's Identity is trivially true since $0 = {n \choose {r-1}} = {n \choose r} = {{n+1} \choose r}$ so we can assume that $r \le n+1$

(4) ${n \choose {r-1}} + {n \choose r} = \dfrac{n!}{(r-1)!(n-r+1)!} + \dfrac{n!}{r!(n-r)!} =$

$$= n!\left(\frac{r}{r!(n-r+1)!}+\frac{n-r+1}{r!(n-r+1)!}\right) = \frac{n!(n+1)}{r!(n-r+1)!} = {{n+1} \choose r}$$