Any prime $p$ between $n + 1$ and $2n + 1$ divides ${2n + 1} \choose {n}$

Question

I don't really know where to begin with this question:

Prove that if $p$ is a prime number such that $n + 1

Answer 1

Hint

Since $$\binom{2n+1}{n}=\frac{(2n+1)(2n)(2n-1) \dotsb (n)}{n!}.$$ and this quantity is an integer. Moreover $p$ appears in the numerator and will not divide anything in the denominator.

Answer 2

$\frac{(2n+1)!}{n!.(n+1)!}$ contains all primes $p$ $>n+1$ and $p \le 2n+1$ as a factor because the denominator can't contain $p$ as a factor and hence can't stop $p$ from being a factor of the numerator.