Prove that $\binom{2n}{n} > \prod_{p > n}^{p<2n}{p}$ Good afternoon everyone. I'm confused about this problem. At the book  A Classical Introduction to Modern Number Theory by Ireland and Rosen is written that the next proposition is true:
$$\binom{2n}{n} > \prod_{p > n}^{p<2n}{p}$$
But,
$$\binom{2n}{n} = \frac{(n+1)(n+2)...(2n)}{n!}$$
We can see that $\frac{(n+1)(n+2)...(2n)}{n!}$ is divisible by all primes such that $n < p < 2n$. That implies:
$$\frac{(n+1)(n+2)...(2n)}{n!} = \frac{(\prod_{p > n}^{p<2n}{p})t}{n!}$$
with $\gcd(t, \prod_{p > n}^{p<2n}{p}) = 1$ This implies that prime numbers $p$ of the decomposition of $t$ are smaller than $n$. Thus, there is a integer $r$ such that:
$$n! = t*r \Rightarrow \frac{(\prod_{p > n}^{p<2n}{p})t}{n!}= \frac{(\prod_{p > n}^{p<2n}{p})t}{t*s} = \frac{(\prod_{p > n}^{p<2n}{p})}{s} \leq \prod_{p > n}^{p<2n}{p} $$
This is a contradiction of the proposition.
What is it wrong of my reasoning?
PD: $n$ is an integer.
 A: This made me want to point out Legendre's Theorem, which states that the power of a prime $p$ that divides $n!$ is given by
\begin{equation*}
\nu_{p}(n) = \sum_{i=1}^{\infty} \Bigg\lfloor \frac{n}{p^{i}} \Bigg\rfloor.
\end{equation*}
Using this it is immediate to see that 
\begin{equation*}
\prod_{p > n, \, p \, \mathrm{prime}}^{p < 2n}p
\end{equation*}
divides $(2n)!$, but no prime $p$ with $n < p < 2n$ divides $n!$. Therefore,
\begin{equation*}
\binom{2n}{n} = \frac{(2n)!}{n!n!}
\end{equation*}
is divisible by the above product as well.
A: Proof that $\binom{2n}{n}> \prod_{p>n}^{p<2n}{p}$ for $n\geq2$:
notice $\binom{2n}{n}$ is even since we can pair every subset with its complement. Also notice that every prime $n<p<2n$ divides $\frac{(2n)!}{n!}$ since it divides the numerator and not the denominator.
We conclude $2\prod_{p>n}^{p<2n}$ divides $\binom{2n}{n}$ and therefore $2\prod_{p>n}^{p<2n}\leq \binom{2n}{n}$
A: Note that $t$ is a multiple of $n!$. You asserted, correctly, that the primes that divide $t$ are primes $\le n$. However, their multiplicities in the prime factorization of $t$ may not be the same as their multiplicities in the prime factorization of $n!$, so we cannot conclude that $t$ divides $n!$.
