Basic Number Theory Inequality assuming Bertrand's postulate false for some n. So I am having a hard time doing one problem from George E. Andrews Number Theory book. It is problem $4$ chapter $8$-$3$ if anyone is curious. For it we must assume that Bertrand's Postulate is false for some $n$ (i.e. for some Integer $n$ there is NO prime $p$ such that $n<p<2n$). Also, we should use in our proof one inequality previously postulated. It is the following: 
$${2n\choose n}\leq\ Q_n \leq\ (2n)^{\pi(2n)}$$ where $Q_n$ is the product of all power of primes not exceeding $2n$ (i.e. the product of all $p^r$ where $r$ is such that $p^r \leq\ 2n <p^{r+1} $) and $\pi(2n)$ is the prime counting function for $2n$.
Knowing all this, I need to prove that $${2n\choose n}\leq\ (2n)^{\sqrt {2n}}R_{\frac{2}{3}n} $$ where $R_{\frac{2}{3}n} $ is the product of all primes less than $\frac{2}{3} n $.
So my first thought was to use the needed assumption to see that $\pi(2n) = \pi(n)$ and from the first inequality I would just need $$(2n)^{\pi(n)} \leq\ (2n)^{\sqrt {2n}}R_{\frac{2}{3}n} $$
to be true. But this is just false even with our assumption. Take for example $n = 7$ then you would get $14^6 = 7529536 \gt  14^{\sqrt {14}}(2 \cdot\ 3)= 116565.406$ 
So I am really stuck now because I find that $\frac{2}{3}n$ too specific and I don't know where it might come from for that to be true. 
EDIT: so I was thinking on $\phi (7)$ when I did that calculation, the actual result for $n = 7$ should be $14^4 = 38416$ and that is indeed less than $116565.406$. So to the way might be to prove the last inequality but I still don't know how to put there the $R_ {\frac{2}{3}n}$. And Knowing that primes are more dense than squares that ratio should also be enough to make that relation flip, considering  the desire result and the exponents on both sides
 A: Once you solve the previous problems in the book you will the idea for this one.
We constructed $Q_n$ by considering every prime $p < 2n : p \space | \binom{2n}{n}$ and finding it's exponent in $\binom{2n}{n}$. Then, the product $Q_n$ is created such that $\binom{2n}{n} | Q_n$
Some points to note (solved in the previous problems in the book): 


*

*If $p>\sqrt{2n}$, then the exponent of $p$ in $Q_n \leq 1$ .

*If $ 2n/3<p<n $, then p does not divide $\binom{2n}{n}$

*Every term $p_i ^ {r_i} \leq 2n$ (definition of $Q_n$)


So, we can go from here
$$ \binom{2n}{n} \leq \prod_{ p \leq \sqrt{2n} } 2n * \prod_{ \sqrt{2n} <  p < 2n/3 } p^1 * \prod_{ n < p < 2n} p $$
Note that the term where $ 2n/3 \leq p \leq n $ is not required.
Now as we have no primes from $n$ to $2n$, the third term is 1.
Therefore, we can write:
$$ \binom{2n}{n} \leq \prod_{ p \leq \sqrt{2n} } 2n * \prod_{ \sqrt{2n} <  p < 2n/3 } p^1$$
$$ \binom{2n}{n} \leq \prod_{ p \leq \sqrt{2n} } 2n * \prod_{ 1 <  p < 2n/3 } p^1$$
$$ \binom{2n}{n} \leq 2n^{\sqrt{2n}} * R_{2n/3}$$
