I want to show that every prime power $p^k$ that divides $\binom{2m}{m}$ is smaller than or equal to $2m$.

As a first step, I looked at $$\binom{2m}{m} = \frac{(2m)!}{(m!)^2} = \frac{2m(2m-1) \ldots (m+2)(m+1)}{m!} \, .$$ Here I'm essentially stuck. I can apply the prime factorization to numerator and denominator, then I can cancel and I know that $p^k$ is left over in the numerator. But I cannot conclude $p^k \leq 2m$.

I feel that some vital ingredient is missing here, but I don't know what it is.

(Post edited with respect to the helpful comment.)

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    $\begingroup$ $\binom 21=2$ is an exception $\endgroup$ – Mark Bennet May 27 '14 at 17:41
  • $\begingroup$ Oh, you're right. It's supposed to be "smaller or equal" and $p^k \leq 2m$. I dropped that because I wasn't aware of the exception. Thanks. $\endgroup$ – Amarus May 27 '14 at 18:11

Hint: Let $\alpha_{n, p}$ be the largest natural number such that $p^{\alpha_{n, p}}\mid n!$ for $n\in\mathbb{N}$ and $p$ prime. Then $$\alpha_{n, p}=\sum_{k=1}^{\infty}{\left[\frac{n}{p^k}\right]},$$ where $[x]$ is the integer part of $x$.

  • $\begingroup$ Does that help, @Amarus? $\endgroup$ – Shaun May 27 '14 at 18:46
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    $\begingroup$ I'm still trying to figure it out. We want to count with what multiplicity $p$ occurs in the prime factorization of $n! \,$. $\left\lfloor \frac{n}{p} \right\rfloor$ is the number of elements in $\left\{1, \ldots, n\right\}$ that are divided by $p$ at least once, $\left\lfloor \frac{n}{p^2} \right\rfloor$ the number of such elements that are divided by $p$ at least two times and so on. So we can add those up until $\left\lfloor \frac{n}{p^k} \right\rfloor$ is less than $1$ to obtain the number of occurences of $p$ in the prime factorization of $n!\,$. Is that about right? $\endgroup$ – Amarus May 27 '14 at 20:47
  • $\begingroup$ Yeah, exactly :) $\endgroup$ – Shaun May 27 '14 at 20:50
  • $\begingroup$ (If this answers your question, please don't forget to accept it by clicking the tick $\ddot\smile$) $\endgroup$ – Shaun May 28 '14 at 13:36
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    $\begingroup$ Another hint: consider $$\alpha_{2n, p}.$$ $\endgroup$ – Shaun May 31 '14 at 10:17

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