Why $\prod\limits_{n
Why is for $p$ prime, $\prod\limits_{n<p\le2n}p\le\binom{2n}{n}$
I think induction doesn't work;
the factor from $\prod\limits_{n-1<p\le2n-2}p\quad$ to $\prod\limits_{n<p\le2n}p$ can be $(2n-1)$
and $\binom{2n}{n}/\binom{2n-2}{n-1}=\frac{2n\cdot(2n-1)}{n^2}$
So the RHS doesn't necessarily ''grow'' faster, can you help ?
 A: Look at the definition of the binomial coefficient
$$
\binom{2n}{n} = \frac{(2n)!}{n!^2}
$$
We know that it is an integer. (This is tricky to prove directly, but we know it is true since a binomial coefficient counts the number of ways something can be done. For a direct proof, prime factorization is a powerful tool.)  For any prime $p$ such that $n<p\leq 2n$, we have that $p$ appears as a factor in the numerator, but it's not canceled by anything in the denominator, so it has to survive and be a factor of the result.
So every such $p$ is a factor of the RHS, and therefore the product of all those primes has to be a factor as well.
A: We have from prime number theorem $$\sum_{p\leq x}\log\left(p\right)\sim x$$
 as $x\rightarrow\infty.$
  Hence $$\log\left(\prod_{n<p\leq2n}p\right)=\sum_{n<p\leq2n}\log\left(p\right)\sim n$$
 so $$\prod_{n<p\leq2n}p\sim e^{n}.$$
 Recall that Stirling approximation is, for $n\in\mathbb{N}^{+}$
 $$\sqrt{2\pi}\frac{n^{n+1/2}}{e^{n}}\leq n!\leq e\frac{n^{n+1/2}}{e^{n}}$$
 we have$$\frac{\left(2n\right)!}{\left(n!\right)^{2}}\geq\frac{\sqrt{2\pi}}{e^{2}}\frac{\left(2n\right)^{2n+1/2}}{e^{2n}}\frac{e^{2n}}{n^{2n+1}}=\frac{\sqrt{2\pi}}{e^{2}}2^{2n+1/2}n^{-1/2}\geq e^{n}.$$
