<s>[INST] Understanding how to estimate $\pi(x)$ based on Paul Erdos's proof of Bertrand's Postulate I am reading the 4th Edition of Proofs from the Book. I am not clear on how the proof behind Bertrand's postulate leads to the following statement on page 10 (of my edition):

From (2) one can derive with the same methods that $\prod\limits_{n<p\le 2n}p \ge 2^{\frac{1}{30}n}$ for $n \ge 4000$

Here's (2):

$4^n \le (2n)^{1+\sqrt{2n}} \cdot \prod\limits_{\sqrt{2n}< p \le \frac{2}{3}n} p \cdot \prod\limits_{n < p \le 2n} p$ for $n \ge 3$

After (2) comes the details that for completing the proof.  I understand how this leads to the proof.  I don't understand how this leads to the estimate for the product of primes
Here are the details presented:
Here's (1):

$\prod\limits_{p \le x}p \le 4^{x-1}$ for all real $x \ge 2$ 

Assuming that there are no primes between $n$ and $2n$, applying (1) to (2):

Substituting (1) into (2) we get:
  $$4^n \le (2n)^{1+\sqrt{2n}}4^{\frac{2}{3}n}$$

Then applying $a+1 < 2^a$, the result becomes:

$$2n = (\sqrt[6]{2n})^6 < (\left\lfloor\sqrt[6]{2n}\right\rfloor + 1)^6 < 2^{6\left\lfloor\sqrt[6]{2n}\right\rfloor} \le 2^{6\sqrt[6]{2n}}$$

Then, for $n \ge 50$ ($18 < 2\sqrt{2n}$), combining the last two results gets:

$$2^{2n} \le (2n)^{3(1 + \sqrt{2n})} < 2^{\sqrt[6]{2n}(18+18\sqrt{2n})} < 2^{20\sqrt[6]{2n}\sqrt{2n}} = 2^{20(2n)^{\frac{2}{3}}}$$

While I follow all the details up to this point in the proof (I think), I don't follow how this gets us to:

From (2) one can derive with the same methods that $\prod\limits_{n<p\le 2n}p \ge 2^{\frac{1}{30}n}$ for $n \ge 4000$

If someone could provide details or an explanation about this last result, I would greatly appreciate it.
 [/INST] We can put (1) into (2) and we get$$4^{n}\leq\left(2n\right)^{1+\sqrt{2n}}\underset{\sqrt{2n}<p\leq\frac{2}{3}n}{\prod}p\underset{n<p\leq2n}{\prod}p\leq\left(2n\right)^{1+\sqrt{2n}}4^{\frac{2}{3}n-1}\underset{n<p\leq2n}{\prod}p\,\Longrightarrow$$
 $$\frac{2^{2n-\frac{4}{3}n-\sqrt{2n}+1}}{n^{1+\sqrt{2n}}}\leq\underset{n<p\leq2n}{\prod}p.$$
 Now, if $n$
  is large enough (and $n\geq4000$
  is large enough) holds$$\frac{1}{n^{1+\sqrt{2n}}}>\frac{1}{2^{n/2}}$$
 so$$2^{2n-\frac{4}{3}n-\sqrt{2n}+1-\frac{n}{2}}\leq\underset{n<p\leq2n}{\prod}p$$
 and trivially $$2n-\frac{4}{3}n-\sqrt{2n}+1-\frac{n}{2}=\frac{12n-8n-6\sqrt{2n}+6-3n}{6}=\frac{n-6\sqrt{2n}+6}{6}>\frac{n}{30}$$
 if $n$ large enough, so we have $$2^{n/30}\leq\underset{n<p\leq2n}{\prod}p.$$
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