Prove that $$ \sum_{i=1}^ni = \binom{n+1}{2} $$

Prove that

$$ \sum_{i=1}^ni = \binom{n+1}{2} $$

$$ \sum_{i=1}^n = \frac{n(n+1)}{2} = \frac{(n+1)!}{2!(n+1-2)!} = \binom{n+1}{2} $$

Interesting. The second diagonals of Pascal's triangle appear to be the arithmetic series.