Show that the product sum of a prime, p, divides $\begin{pmatrix}2n + \delta\\n\end{pmatrix}.$

Question

Let $\delta\in \{0,1\}$ be fixed. Show that if a prime $p$ that satisfies $n + \delta < p \leq 2n + \delta$, then $p$ is a divisor of $$\begin{pmatrix}2n + \delta\\n\end{pmatrix}.$$ Use this to show that

$$\prod_{ n + \delta < p \leq 2n + \delta } p \mid \begin{pmatrix}2n + \delta\\n\end{pmatrix}$$ where the product $p$ is of all primes $p$ in the interval $\{n + 1 + \delta,\ldots, 2n + \delta\}$

Im completely clueless on how to begin this problem. Any hints or advice would be appreciated.

Answer 1

Write out $2n+\delta \choose n$ using factorials, and note that $p$ divides the numerator but not the denominator.