Here are some simple proofs based on the binomial theorem. They also demonstrate how amazing mathematical proofs can be.
1. The hockey stick identity
By the formula for the sum of a geometric series, \(\frac{(1+r)^{n+1}-1}{(1+r)-1}=\displaystyle\sum_{i=0}^n (1+r)^i\). Now expanding both sides by the binomial theorem and simplifying, $$\displaystyle\sum_{i=0}^n \binom{n+1}{i+1}r^i=\displaystyle\sum_{i=0}^n \displaystyle\sum_{j=0}^i \binom{i}{j}r^j$$ Now rearrange the sum on the left (it is a finite sum after all) to get $$\displaystyle\sum_{i=0}^n \binom{n+1}{i+1}r^i=\displaystyle\sum_{j=0}^n r^j\displaystyle\sum_{i=j}^n \binom{i}{j}$$ Equating coefficients of \(r^k\) on both side gives the dsired result.
2. Vandermonde’s identity
Well, obviously, \((1+r)^n (1+r)^m =(1+r)^{n+m}\). Again, use the binomial theorem to expand both sides: $$\left(\displaystyle\sum_{i=0}^n \binom{n}{i}r^i \right)\left(\displaystyle\sum_{i=0}^m \binom{m}{i}r^i \right)=\displaystyle\sum_{i=0}^{n+m} \binom{n+m}{i}r^i$$ The coefficient of \(r^k\) in the LHS is $$\displaystyle\sum_{i+j=k}\binom{n}{i}\binom{m}{j} =\displaystyle\sum_{i=0}^k \binom{n}{i}\binom{m}{k-i}$$ so equating coeffiecients of \(r^k\) yields Vandermonde’s identity.
3. Tri-Vandermonde
Same as #2, except with the product of three binomials.