Factorial Proof - ${n \choose r-1}+{n \choose r}={n+1 \choose r}$ ${n \choose r-1}+{n \choose r}={n+1 \choose r}.$
So what I tried to do was expand the first and second term.
$\frac{n!}{(r-1)!(n+1-r)!}+\frac{n!}{(r)!(n-r)!}.$
Then what I did was try to get common denominators.
$\frac{n!}{(r-1)!(n+1-r)(n-r)!}+\frac{n!}{(r)(r-1)!(n-r)!}.$
Then I attempted to combine and get the common denominator.
$\frac{(n!)(r)+(n!)(n+1-r)}{(r-1)!(n+1-r)(n-r)!(r)}.$
From here I simplified a bit more but didn't get a nice answer.
Some insight would be helpful. 

You start well:
$$
\binom{n}{r-1}+\binom{n}{r}=
\frac{n!}{(r-1)!\,(n-r+1)!}+\frac{n!}{r!\,(n-r)!}
$$
Now it's best to collect all common terms. Not so different from what you did, but with less complications. Note you can collect


*

*$n!$ in the numerator

*$(r-1)!$ in the denominator, using $r!=r\,(r-1)!$

*$(n-r)!$ in the denominator, using $(n-r+1)!=(n-r+1)\,(n-r)!$
Thus you can go on with
$$
=\frac{n!}{(r-1)!\,(n-r)!}\left(\frac{1}{n-r+1}+\frac{1}{r}\right)
$$
The part in parentheses can be rewritten as $\dfrac{n+1}{r(n-r+1)}$
so you have
$$
=\frac{(n+1)\,n!}{r\,(r-1)!\,(n-r+1)\,(n-r)!}=\frac{(n+1)!}{r!\,(n+1-r)!}=\binom{n+1}{r}
$$
