A Double Sum of Binomials

Prove the following summation of binomial coefficients.
$\sum_{i=0}^n \sum_{j=0}^m {i+j \choose j} = {n + m +2 \choose n+1} – 1.$

First, by Pascal’s rule that $${i+j \choose j} = {i+j-1 \choose j} + {i+j-1 \choose j-1}$$, the inner summation
$\sum_{j=0}^m {i+j \choose j} = {i+m+1 \choose m}.$

Similarly,
$\sum_{i=0}^n {i+m+1 \choose m} = \sum_{i=0}^n {i+m+1 \choose i+1} = \sum_{i=0}^{n+1} {i+m+1 \choose i} – 1 = {n+m+2 \choose n+1} – 1.$