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
$\binom{2n}{n}$. 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 :-)