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  "body": "\n## Proof\n\nWe argue by induction on $n$. The claim for $n$ is that the identity holds for this $n$, every natural number $m$ and every natural number $r$.\n\n1. Let $n = 0$. By the definition of the [[binomial coefficient|092GEF38YD]], $\\binom{0}{0} = 1$ and $\\binom{0}{j} = 0$ for $j \\ge 1$.\n2. So in $\\sum_{k=0}^{r} \\binom{m}{k} \\binom{0}{r-k}$ only the term $k = r$ is nonzero. The sum equals $\\binom{m}{r} = \\binom{m+0}{r}$.\n3. Suppose the claim holds for $n$. Fix $m$ and $r$. We prove $\\binom{m+n+1}{r} = \\sum_{k=0}^{r} \\binom{m}{k} \\binom{n+1}{r-k}$.\n4. If $r = 0$, both sides equal $1$, because $\\binom{a}{0} = 1$ for every $a$.\n5. Otherwise $r = s + 1$ for a natural number $s$. By [[Pascal's rule|0YRG241HQQ]], $\\binom{m+n+1}{s+1} = \\binom{m+n}{s} + \\binom{m+n}{s+1}$.\n6. By the induction hypothesis for $n$, with $r = s$ and with $r = s + 1$, the left side is $L_1 + L_2$. Here $L_1 = \\sum_{k=0}^{s} \\binom{m}{k} \\binom{n}{s-k}$ and $L_2 = \\sum_{k=0}^{s+1} \\binom{m}{k} \\binom{n}{s+1-k}$.\n7. In the right side $\\sum_{k=0}^{s+1} \\binom{m}{k} \\binom{n+1}{s+1-k}$, the term $k = s + 1$ is $\\binom{m}{s+1} \\binom{n+1}{0} = \\binom{m}{s+1} \\binom{n}{0}$, since both coefficients $\\binom{\\cdot}{0}$ equal $1$.\n8. For $k \\le s$, write $s + 1 - k = (s - k) + 1$. Pascal's rule gives $\\binom{n+1}{s+1-k} = \\binom{n}{s-k} + \\binom{n}{s+1-k}$.\n9. So the right side equals $\\sum_{k=0}^{s} \\binom{m}{k} \\binom{n}{s-k} + \\sum_{k=0}^{s} \\binom{m}{k} \\binom{n}{s+1-k} + \\binom{m}{s+1} \\binom{n}{0}$.\n10. The first sum is $L_1$. The last two parts together are $L_2$, whose term $k = s + 1$ is $\\binom{m}{s+1} \\binom{n}{0}$.\n11. Hence the right side equals $L_1 + L_2$, which is the left side by step 6. This completes the induction.\n\n## Idea\n\nPascal's rule splits every binomial coefficient with top entry $n + 1$ into two with top entry $n$.\nApplying it on both sides reduces the identity for $n + 1$ to two instances of the identity for $n$.\nThe induction hypothesis must cover all $r$, because the step uses it for $r - 1$ and for $r$.\n",
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