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  "body": "\n## Proof\n\nWe use the recursive definition of the [[binomial coefficient|092GEF38YD]] and the [[factorial|0WFCGVMRRC]], and argue by induction on $n$. We prove the division-free form $k!\\,(n-k)!\\,\\binom{n}{k} = n!$.\n\n1. Lemma: for every natural number $n$, $\\binom{n}{n} = 1$, and $\\binom{n}{j} = 0$ for all $j \u003e n$. We prove it by induction on $n$.\n2. For $n = 0$, the definition gives $\\binom{0}{0} = 1$ and $\\binom{0}{j} = 0$ for $j \\ge 1$.\n3. Suppose the lemma holds for $n$. Let $j \u003e n + 1$, and write $j = i + 1$ with $i \u003e n$. By [[Pascal's rule|0YRG241HQQ]], $\\binom{n+1}{i+1} = \\binom{n}{i} + \\binom{n}{i+1} = 0 + 0 = 0$.\n4. Also $\\binom{n+1}{n+1} = \\binom{n}{n} + \\binom{n}{n+1} = 1 + 0 = 1$. This proves the lemma.\n5. Main claim: for every $n$ and every $k \\le n$, $k!\\,(n-k)!\\,\\binom{n}{k} = n!$. We argue by induction on $n$.\n6. For $n = 0$, only $k = 0$ occurs, and $0!\\,0!\\,\\binom{0}{0} = 1 = 0!$.\n7. Suppose the claim holds for $n$, and let $k \\le n + 1$. If $k = 0$, then $0!\\,(n+1)!\\,\\binom{n+1}{0} = (n+1)!$, since $\\binom{n+1}{0} = 1$.\n8. If $k = n + 1$, then $(n+1)!\\,0!\\,\\binom{n+1}{n+1} = (n+1)!$ by the lemma.\n9. Otherwise $1 \\le k \\le n$. Write $k = j + 1$ with $0 \\le j \\le n - 1$. Then $n + 1 - k = n - j$, and Pascal's rule gives $\\binom{n+1}{k} = \\binom{n}{j} + \\binom{n}{j+1}$.\n10. For the first term, $(j+1)! = (j+1) \\cdot j!$. By the induction hypothesis for $j$, $(j+1)!\\,(n-j)!\\,\\binom{n}{j} = (j+1) \\cdot n!$.\n11. For the second term, $n - j \\ge 1$, so $(n-j)! = (n-j) \\cdot (n-j-1)!$. By the induction hypothesis for $j + 1 \\le n$, $(j+1)!\\,(n-j)!\\,\\binom{n}{j+1} = (n-j) \\cdot n!$.\n12. Adding steps 10 and 11 gives $k!\\,(n+1-k)!\\,\\binom{n+1}{k} = (j + 1 + n - j) \\cdot n! = (n+1) \\cdot n! = (n+1)!$.\n13. This completes the induction. So $k!\\,(n-k)!\\,\\binom{n}{k} = n!$ whenever $k \\le n$.\n14. Every factorial is at least $1$, since $0! = 1$ and $(m+1)! = (m+1) \\cdot m!$. So $k!\\,(n-k)! \\ne 0$.\n15. Dividing by $k!\\,(n-k)!$ gives $\\binom{n}{k} = \\dfrac{n!}{k!\\,(n-k)!}$.\n\n## Idea\n\nBoth sides obey the same recursion. Pascal's rule splits $\\binom{n+1}{k}$ into two terms.\nAfter multiplying by $k!\\,(n+1-k)!$, the two terms become $k \\cdot n!$ and $(n+1-k) \\cdot n!$, and these add up to $(n+1)!$.\n",
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