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I can use either combinatorial argument, or mathematical induction on ${n+1\choose k} = {n\choose k}+{n\choose k-1}$ to prove $\prod_{i=0}^{k-1}(n-i)$ being divisible by $k!$ or that $n \choose k$ is a natural number for natural numbers $n$ and $k$. What is a number theoretical proof for it that does not use mathematical induction? Please do not give me a combinatorial argument with which I am infinitely familiar!

Bill Dubuque
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Hans
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    Worth pointing out... if you were to take $\binom{n}{k}$ with the combinatorial or set theoretic definition instead ($\binom{n}{k}$ being the number of $k$-element subsets of an $n$-element set), that it is a natural number is obvious since it is the answer of a counting problem and all counting problems necessarily have non-negative integer answers. – JMoravitz Jun 07 '21 at 15:25
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    Just compare the prime divisors of n! and (n-k)!k! – LurchiDerLurch Jun 07 '21 at 15:26
  • @Hans hence why it is a comment, not an answer. That said, there are a number of things which may start as algebraic objects who can be given a combinatorial interpretation, allowing you to use combinatorial arguments to conclude things about them and this can be far simpler and more intuitive to use than direct arguments. Do not discount the usefulness of trivial arguments. – JMoravitz Jun 07 '21 at 16:03
  • you can prove pascal's identity purely algebraically... – C Squared Jun 07 '21 at 16:24
  • @CSquared: Please read the question carefully. – Hans Jun 07 '21 at 16:25
  • you said you can prove pascals identity by mathematical induction or combinatorically. never said anything about how you proved $\prod_{i=0}^{k-1}(n-i)$ is divisible by $k!$, other than it relying on your non-algebraic proof of pascals identity – C Squared Jun 07 '21 at 16:28
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    Here is a simple algorithm that proves binomial coefficients are integers by rewriting their fraction as a product of fractions whose denominators are coprime to any given prime. But that uses induction (recursion) as does almost any proof that a (nontrivial) statement holds true for all integers. – Bill Dubuque Jun 07 '21 at 17:16

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A number-theoretic proof can be based on looking at the multiplicity of each prime number $p$ in the rational number $\binom{n}{k}$: prove the multiplicity is nonnegative for every $p$. See the proof in Section 5 of the file here. Other proofs are there too, all of which can be extended to prove $q$-binomial coefficients are polynomials in $q$ with integral coefficients; ordinary binomial coefficients are the special case $q=1$.

KCd
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  • Thank you. The number theoretic proof of your reference invokes Legendre's formula for $m_p(N!)$. Do you have an English reference for its proof? – Hans Jun 07 '21 at 16:40
  • @Hans it's trivial to find a proof of Legendre's formula online by googling for it. – KCd Jun 07 '21 at 18:22
  • +1 and accepted. Actually, come to think of it, Legendre's formula is easy to understand and prove. – Hans Jun 07 '21 at 18:37
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    The reason I previously edited your answer to show the author and title of the paper you referenced was often links would break which would loose completely the information invoked whereas people can still google the paper had the author and title been provided. Please kindly consider this point. – Hans Jun 07 '21 at 22:32