Show that if $p$ is a prime and $p \in (n, 2n]$, then $p \mid {2n \choose n}$.

Question

I'm having a problem understanding the answer to this question below. The step I don't get is underlined in red. I understand everything else just the red underline I am stuck on. Sorry I am a beginner so probably it is something very simple but I can't see it .

Question :

Answer :

Answer 1

Claim $1$: If $a \mid c$, $b \mid c$ and $\gcd(a,b)=1$, then $ab \mid c$.

Now note that every prime $p$ between $n$ and $2n$ divides $\dbinom{2n}n$. From Claim $1$, it follows that $\prod_{n < p < 2n}p$ divides $\dbinom{2n}n$.

Answer 2

Call $P_n= \{ p \mbox{ prime} : n < p \le 2n\}$. You just showed that for all $p \in P_n$ you have that $p$ divides $\binom{2n}{n}$. So in the prime factorization of $\binom{2n}{n}$ all such primes appear.

In particular $\prod_{p \in P_n} p$ divides $\binom{2n}{n}$, hence it is smaller.