Asymptotics of the n-th prime using the gamma function

The asymptotic expansion of Cipolla starts $$p_n=n\log n+n\log\log n-n+n\frac{\log\log n}{\log n}+O(n(\log\log n/\log n)^2)$$ So the given approximations have errors $$p_n=n\frac{\Gamma'(n)}{\Gamma(n)}+\Theta(n\log\log n)$$ and $$p_n=n\log\frac{\Gamma'(n)}{\Gamma(n-1)}+\Theta(n).$$ I would not say these are good approximations with so big errors.

The inverse function of the log integral function $\text{li}^{-1}(x)$ has error $$p_n= \text{li}^{-1}(n) +O(n \exp(-c\sqrt{\log n})$$ which assumming Riemann hypothesis can be reduced to $$|p_n-\text{li}^{-1}(n)|\le \pi^{-1} \sqrt{n}(\log n)^{\frac52}\qquad n>11.$$ (see arXiv:1203.5413)

Can someone explain why the gamma function approximated the n-th prime so nicely?
Is this a coincidence or is there some underlying phenomenon governing this result?

The expression which appears there is $\dfrac{\Gamma'(n)}{\Gamma(n)}=\Big[\ln\Gamma(n)\Big]'=\psi_0(n)$, see digamma function.

This famous function has the property that $\psi_0(n)=H_{n-1}-\gamma$, where $H_n$ is the n-th harmonic

number, and $\gamma\simeq\dfrac1{\sqrt3}$ is the Euler-Mascheroni constant. But, at the same time, $H_{n-1}\simeq\ln n$,

the small error or difference between the two tending towards the afore-mentioned constant. By

differentiating the natural logarithm of the fundamental property of the $\Gamma$ function, $\dfrac{\Gamma(n+1)}{\Gamma(n)}=n$,

we have that $\psi_0(n+1)=\dfrac1n+\psi_0(n)$, which is nearly identical to the recurrence relation of

harmonic numbers, $H_{n+1}=\dfrac1{n+1}+H_n$. The only difference is the offset, $\psi_0(1)=-\gamma\neq$

$\neq H_0=0$, a proof of which can be found here. The conclusion then follows from the prime number theorem.