I have a very hard proof from "Proofs from the BOOK". It's the section about Bertrand's postulate, page 9:
I have to show, that for $\frac{2}{3}n there is no p which divides $\binom{2n}{n}$. I know $\binom{2n}{n}=\frac{(2n)!}{n!n!}$ and from $\frac{2}{3}n I follow $3p>2n$. Then $(2n)!$ has only the prime factors $p$ and $2p$ (because 3p>2n) and $n!$ has only the prime factor $p$. At this point the author goes on to the next part of the proof. Can someone explain to me, how the argument about the p's proofs the statement, that there is no $p$ which divides $\binom{2n}{n}$? I hope my question is clear and sorry for my bad English. Thanks in advance :-)