Combinatorial proof of a binomial identity

You could try using proof by induction:
You need this formula, $\binom{n}{j}=\frac{n!}{j!\left(n-j\right)!}$

1: let r=1 and show that the two sides are equal.

2: Assume that it's true for r = k.

This tells us, $\sum_{i=0}^{k}\binom{n+i}{n}=\binom{n+k+1}{n+1}$

3: Now prove that it's true for r = k+1
$\sum_{i=0}^{k+1}\binom{n+i}{n}=\sum_{i=0}^{k}\bino m{n+i}{n}+\binom{n+k+1}{n}$

This is equal to, $\binom{n+k+1}{n+1}+\binom{n+k+1}{n}=\binom{n+k+2}{ n+1}$

Now making use of the formula from the start

$\frac{\left(n+k+1\right)!}{\left(n+1\right)!\left( k\right)!} +\frac{\left(n+k+1\right)!}{n!\left(k+1\right)!} = \frac{\left(n+k+2\right)!}{\left(n+1\right)!\left( k+1\right)!}$

simplify the LHS

$\frac{\left(n+k+1\right)!n!\left(k+1\right)! +\left(n+k+1\right)!\left(n+1\right)!k!}{\left(n+1 \right)!\left(k!\right)\left(n\right)!\left(k+1\ri ght)!}= \frac{\left(n+k+2\right)!}{\left(n+1\right)!\left( k+1\right)!}$

using $n! = n\left(n-1\right)!$ and simplifying further

$\frac{\left(n+k+1\right)!n!\left(k+1\right)\left(k \right)! +\left(n+k+1\right)!\left(n+1\right)\left(n\right) !k!}{\left(n+1\right)!\left(k!\right)\left(n\right )!\left(k+1\right)!}= \frac{\left(n+k+2\right)\left(n+k+1\right)!}{\left (n+1\right)!\left(k+1\right)!}$

You can show the two sides are equal for k+1 so equal for all k ( or r)