You may find the following result interesting. The point is to highlight that by reasoning in terms of $p_n -n $ it is possible to find new expressions of the asymptotic trend.

let $p_{n}$ be the nth-prime number and $W_{t}$ the Lambert-W function / $W(z)e^{W(z)}=z $ /:

$ \exists \ \ l \in \mathbb N:\forall n>l$ 
 
$$ p_{n}<n-nW_{-1}\bigg(\frac{-e^{2}}{n}\bigg)$$

C. AXLER in [1] showed that $ \exists \ \ l \in \mathbb N:\forall n>l$ :

>$ p_{n}<n\Big(\ln(n)+\ln_{2}(n)-1+\frac{\ln_{2}(n)-2}{\ln(n)}-\frac{\ln_{2}(n)^{2}-6\ln_{2}(n)+10.667}{2\ln(n)^{2}}\Big)=\Theta_{n}$ 
$ p_{n}>n\Big(\ln(n)+\ln_{2}(n)-1+\frac{\ln_{2}(n)-2}{\ln(n)}-\frac{\ln_{2}(n)^{2}-6\ln_{2}(n)+10.667}{2\ln(n)^{2}}\Big)=\Phi_{n}  $ 
>

So

$p_{n}=n\Big(\ln(p_{n}-n)-\ln(p_{n}-n)+\frac{p_{n}}{n}\Big)<n\Big(\ln(p_{n}-n)-\ln(\Phi_{n} -n)+\frac{\Theta_{n}}{n}\Big)$

$j_n=\ln(\Phi_{n} -n)-\frac{\Theta_{n}}{n}$

Since we have $j_n>1 \ $for $n\geq32$ (by computation)

$\ln(p_{n}-n)+\frac{p_{n}}{n}>j_n>1$



$p_{n}<n\Big(\ln(p_{n}-n)-j_n\Big)<n\Big(\ln(p_{n}-n)-1\Big)$.


>$p_n+n<n\ln(p_{n}-n)$

$e^{\ \frac{p_n}{n}+1}<p_n-n$

$e^{2}e^{\ \frac{p_n}{n}-1}<n(\frac{p_n}{n}-1)$

$-\frac{e^{2}}{n}>-(\frac{p_n}{n}-1) e^{-(\ \frac{p_n}{n}-1)}$

>$$ p_{n}<n-nW_{-1}\bigg(\frac{-e^{2}}{n}\bigg) $$




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[1] NEW ESTIMATES FOR THE n-TH PRIME NUMBER, CHRISTIAN AXLER, Jun 2017 https://arxiv.org/pdf/1706.03651.pdf


https://math.stackexchange.com/questions/3476635/prime-numbers-upper-bound-involving-lambert-w-proof

https://www.researchgate.net/publication/258373882_Primes_and_the_Lambert_W_function