Proving that there are at least $n$ primes between $n$ and $n^2$ for $n \ge 6$

Question

I was thinking about Paul Erdos's proof for Bertrand's Postulate and I wondered if the basic argument could be used to show that there are more than $n$ primes between $n$ and $n^2$.

Is this approach valid? Is there a better approach?

Here's the argument:

1. Let $v_p(n)$ be the maximum power of $p$ that divides $n$.
2. Using Legendre's Theorem, I calculate that $v_p\left(\dfrac{(n^2)!}{(n!)^n}\right) \le (n-1)n^2$.
   
   
   1. Let $n = qp^i + r$ where $0 \le r < p^i$
   2. For $r < \dfrac{p^i}{n}$, $\left(\left\lfloor\dfrac{n^2}{p^i}\right\rfloor - n\cdot\left\lfloor\dfrac{n}{p^i}\right\rfloor\right) = 0$
   3. For $r \ge \dfrac{p^i}{n}$, $\left(\left\lfloor\dfrac{n^2}{p^i}\right\rfloor - n\cdot\left\lfloor\dfrac{n}{p^i}\right\rfloor\right) \le n-1$
   4. For $p^i > n^2$, $\left(\left\lfloor\dfrac{n^2}{p^i}\right\rfloor - n\cdot\left\lfloor\dfrac{n}{p^i}\right\rfloor\right) = 0$
3. Using the multinomial theorem I get: $\dfrac{n^{n^2}}{n^2+1} < \dfrac{(n^2)!}{(n!)^n}$.
   1. $n^{n^2}=(1 + 1 + \cdots + 1)^{n^2} = \sum\limits_{k_1+k_2+\cdots+k_n=n^2} {n^2 \choose k_1, k_2, \ldots, k_n} \prod\limits_{1\le t\le n^2}x_{t}^{k_{t}}\,,$ where ${n^2 \choose k_1, k_2, \ldots, k_n} = \frac{(n^2)!}{k_1!\, k_2! \cdots k_n!}$
   2. $n^{n^2} < (n^2+1)\left(\dfrac{(n^2)!}{(n!)^n}\right)$
   3. $\dfrac{n^{n^2}}{n^2+1} < \dfrac{(n^2)!}{(n!)^n}$
4. $\dfrac{(n^2)!}{(n!)^n} < [(n-1)(n^2)]^n\prod\limits_{n < p \le n^2} n$
5. $\dfrac{n^{n^2}}{n^2+1} < [(n-1)(n^2)]^n\prod\limits_{n < p \le n^2} n$
6. For $n \ge 1, \dfrac{n^{n^2}}{n^{4n}} < \dfrac{n^{n^2}}{(n^2+1)[(n-1)(n^2)]^n} < \prod\limits_{n < p \le n^2} p$
7. For $n \ge 6$, $n^2 - 4n = n(n-2) \ge 2n$
8. $(2\log n)(\pi(n^2) - \pi(n)) > \sum\limits_{n < p \le n^2} \log p > (n^2-4n)\log n > (2n)\log n$
9. So $\pi(n^2) - \pi(n) > n$.

Thanks,

-Larry

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Edit: I believe that step 4 is not correct. If I can save the argument, I will update it. Otherwise, please check out the link in the answer.

Answer 1

We have the following result¹:

(Chebyshev, 1850) For any $X>30$, we have: $$\pi(X)\cdot\frac{\log X}{X} \in (A,B),$$ where $A=\log\left(\frac{2^{\frac{1}{2}}3^{\frac{1}{3}}4^{\frac{1}{4}}}{30^{\frac{1}{30}}}\right)\approx 0.946$ and $B=\frac{6}{5}A\approx 1.135$.

It follows that for any $X>30$ we have: $$ \pi(X^2)-\pi(X)\geq A\frac{X^2}{2\log X}-B\frac{X}{\log X}\geq\left(\frac{A}{2}-\frac{B}{30}\right)\frac{X^2}{\log X}\geq\color{red}{\frac{2}{5}\frac{X^2}{\log X}} $$ that is way greater than $X$. So we just need to check that the our claim holds in the range $[6,30]$.

^{1) it is a weaker version of the PNT, but not so weaker. Its proof only relies on the properties of the central binomial coefficients, and it is rather short and ingenious.}