TITLE: On a property of the binomial coefficient
QUESTION [5 upvotes]: Let $n$ be a positive integer and $p$ a prime, how can I prove that the highest power of $p$ that divides $\binom{2n}{n}$ is exactly the number of $k\geq1$ such that $\lfloor 2n/p^k\rfloor$ is odd?

REPLY [2 votes]: HINT:
This is a useful general approach to questions regarding the divisibility of factorials and binomials. What is the highest power of $p$ that divides $n!$? Well $p, 2p, 3p,...$ and all further multiples up to $n$ divide $n!$, so at least $ \left\lfloor \frac{n}{p} \right\rfloor $. But that was not counting $2$ powers for every square, $p^2, 2p^2...$. So we need to tack on another $\left\lfloor \frac{n}{p^2} \right\rfloor$. And now we need to account for cubes and so on, so the highest power of $p$ that divides $n!$ is $$ \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor. $$
Now, since $ \binom{2n}{n} = \frac{(2n)!}{(n!)^2} $, the highest power of $p$ to divide that is given by: $$ \sum_{k=1}\left( \left\lfloor \frac{2n}{p^k} \right\rfloor - 2\left\lfloor \frac{n}{p^k}\right\rfloor \right) $$
Investigate the general term in cases and deduce your result.
Edit: Thijs, your recent edit ended up being essentially my post.