Show that $p_n^{1-\epsilon}\le n$ using PNT Assuming PNT $$\pi(x)\sim \frac{x}{\log{x}}$$
How can we show that given any $\epsilon>0$ $$p_n^{1-\epsilon}< n,$$ for all sufficiently large $n$ ($p_n$ denotes the $n^{th}$ prime.)
My work:
Setting $x=p_n$ we get $\lim\limits_{n\to\infty}\dfrac{n\log{p_n}}{p_n}=1\Rightarrow 1-\epsilon<\dfrac{n\log{p_n}}{p_n}$, for sufficiently large $n$. Now if I can show that $$\dfrac{n\log{p_n}}{p_n}\le\dfrac{\log n}{\log{p_n}}$$
then the result follows. But I am unable to show the last inequality.
Can someone please help with this? Other approaches are welcome as well.
[NOTE: Using this I want to show that $\lim\limits_{n\to\infty}\frac{\log{n}}{\log{p_n}}=1$ so please don't use that, although it can be derived independently]
Thank you
 A: $\color{red}{\text{1. One way}}$
I will use 2 results


*

*PNT
$$\lim\limits_{n\rightarrow\infty}\frac{\pi(n)\ln{(n)}}{n}=1 \Rightarrow
\lim\limits_{n\rightarrow\infty}\frac{n}{\pi(n)\ln{(n)}}=1 \tag{1}$$

*and
$$\lim\limits_{n\rightarrow\infty}\frac{p_n}{n\ln{(n)}}=1 \tag{2}$$



Proposition 1.1 $$\lim\limits_{n\rightarrow\infty} \frac{\ln{(n)}}{\ln{(p_n)}}=1$$

$\{p_n\}$ is a subsequence of $\{n\}$, thus, from $(1)$, 
$$\lim\limits_{p_n\rightarrow\infty}\frac{\pi(p_n)\ln{(p_n)}}{p_n}=1 \Rightarrow 
\lim\limits_{n\rightarrow\infty}\frac{\pi(p_n)\ln{(p_n)}}{p_n}=1 \Rightarrow ...$$
because $\pi(p_n)=n$
$$...\lim\limits_{n\rightarrow\infty}\frac{n\ln{(p_n)}}{p_n}=1 \tag{3}$$
Now
$$\lim\limits_{n\rightarrow\infty} \frac{\ln{(n)}}{\ln{(p_n)}}=
\lim\limits_{n\rightarrow\infty} \left(\frac{n\ln{(n)}}{p_n}\cdot\frac{p_n}{n\ln{(p_n)}}\right)=\\
\lim\limits_{n\rightarrow\infty} \left(\frac{n\ln{(n)}}{p_n}\right)\cdot \lim\limits_{n\rightarrow\infty} \left(\frac{p_n}{n\ln{(p_n)}}\right)\overset{(2)(3)}{=}1$$

Proposition 1.2 For large enough $n$  $$p_n^{1-\varepsilon}<n$$

From 
$$\lim\limits_{n\rightarrow\infty} \frac{\ln{(n)}}{\ln{(p_n)}}=1$$
using the definition of limit, $\forall\varepsilon >0, \exists N(\varepsilon)\in\mathbb{N}$ s.t. $\forall n> N(\varepsilon)$
$$\left|\frac{\ln{(n)}}{\ln{(p_n)}}-1\right|<\varepsilon \Rightarrow
1-\varepsilon <\frac{\ln{(n)}}{\ln{(p_n)}}< 1+\varepsilon \Rightarrow \\
(1-\varepsilon)\ln{(p_n)} <\ln{(n)}< (1+\varepsilon)\ln{(p_n)} \Rightarrow \\
\ln{(p_n)^{(1-\varepsilon)}} <\ln{(n)}< \ln{(p_n)^{(1+\varepsilon)}} \Rightarrow ...$$
$e^x$ is ascendng, thus
$$... p_n^{1-\varepsilon} <n< p_n^{1+\varepsilon} $$

$\color{red}{\text{2. Another way}}$
Using Vallée-Poussin, for large enough $x$
$$\pi(x)>\frac{x}{\ln(x)-(1-\varepsilon)}>\frac{x}{\ln(x)}$$
Let's show that for large enough $x$ we also have 
$$\frac{x}{\ln(x)}>x^{1-\varepsilon}$$
which is the same as showing $$\frac{x^{\varepsilon}}{\ln{x}}>1$$
for large $x>0$.

Propositions 2.1 Function $f(x)=\frac{x^{\varepsilon}}{\ln{x}}$ is ascending for large $x>0$.

Because
$$f'(x)=\frac{x^{\varepsilon-1} (\varepsilon \ln{x}-1)}{\ln^2{x}}>0 \iff 
\varepsilon \ln{x}-1>0 \Rightarrow 
x> e^{\frac{1}{\varepsilon}}$$

Propositions 2.2 $\lim\limits_{x\rightarrow \infty}f(x) \rightarrow \infty$

If we assume it is bounded by a large $\alpha>0, \forall x>1$ and we know $\ln{x}$ is ascending
$$\frac{x^{\varepsilon}}{\ln{x}} < \alpha \iff 
1<x^{\varepsilon}< \alpha \ln{x} \iff 
\color{red}{0}<\varepsilon<\frac{\ln{\alpha}}{\ln{x}}+\frac{\ln{\ln{x}}}{\ln{x}}\rightarrow \color{red}{0}, x\rightarrow\infty$$
which is a contradiction.
So, for large $x$ we have 
$$\pi(x)>x^{1-\varepsilon}$$
which means, for large $n$ we have
$$n=\pi(p_n)>p_n^{1-\varepsilon}$$
