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  "body": "\n## Proof\n\nFix $r$. We argue by induction on $n \\ge r$.\n\n1. Let $n = r$. The left side is the single term $\\binom{r}{r}$.\n2. By the [[factorial formula|033KTKKRV6]], $\\binom{r}{r} = \\frac{r!}{r!\\,0!} = 1$ and $\\binom{r+1}{r+1} = 1$. So both sides equal $1$.\n3. Suppose $\\sum_{i=r}^{n} \\binom{i}{r} = \\binom{n+1}{r+1}$ for some $n \\ge r$.\n4. Then $\\sum_{i=r}^{n+1} \\binom{i}{r} = \\binom{n+1}{r+1} + \\binom{n+1}{r}$, by adding the term $i = n + 1$.\n5. By [[Pascal's rule|0YRG241HQQ]] with $n + 1$ and $r$, $\\binom{n+1}{r} + \\binom{n+1}{r+1} = \\binom{n+2}{r+1}$.\n6. So $\\sum_{i=r}^{n+1} \\binom{i}{r} = \\binom{(n+1)+1}{r+1}$. This completes the induction.\n\n## Idea\n\nAdding one more term to the diagonal sum adds $\\binom{n+1}{r}$ to $\\binom{n+1}{r+1}$.\nPascal's rule turns this pair into the next entry $\\binom{n+2}{r+1}$, so the formula moves one row down.\n",
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