Regularization involving Stieltjes constants: $\displaystyle\sum_{k=1}^{\infty}\frac{\ln(k)^n}{k}\overset{\mathcal{R}}{=}\gamma_n$

Question

Notation

• $\zeta(z)$ is the Riemann zeta function
• $\operatorname{Li}_{\nu}(z)$ is the polylogarithm function
• $\operatorname{Li}^{(n,0)}_{\nu}(z):=\frac{\partial^n}{\partial\nu^n}\operatorname{Li}_\nu(z)$
• $\gamma_n$ is the $n$-th Stieltjes constant
• $\overset{\mathcal{R}}{=}$ means "equal in the sense of regularizations"

Context

Starting from the fact that, for $s\to 0$: $$\zeta(s+1)=\frac{1}{s}+\sum_{k=0}^{\infty}(-1)^k\gamma_k\frac{s^k}{k!}$$ $$\operatorname{Li}_{s+1}(z)=\sum_{k=0}^{\infty}\operatorname{Li}_1^{(k,0)}(z)\frac{s^k}{k!}$$ And $\operatorname{Li}_{s+1}(1)=\zeta(s+1)$

We come to deduce that $$\operatorname{Li}_1^{(n,0)}(1)\overset{\mathcal{R}}{=}(-1)^n\gamma_n$$ From which, explaining the form of the polylogarithm, we obtain: $$\sum_{k=1}^{\infty}\frac{\ln(k)^n}{k}\overset{\mathcal{R}}{=}\gamma_n$$

Question

I would like to know if these steps make sense from the point of view of regularizations (or generally make sense in some kind of non-standard analysis) or is it a coincidence. Because on Wolfram I found that: $$\gamma_n=\lim_{m\to\infty}\left(\sum_{k=1}^{\infty}\frac{\ln(k)^n}{k}-\frac{\ln(m)^{n+1}}{n+1}\right)$$ Which is a very similar form to the previous one. Is it a coincidence or is there some connection? Is it possible to demonstrate this more correctly?