How do we show that $\int_{0}^{1}{\ln^k(x)\over x}\ln\left(1-\sqrt[n]{x}\right)\mathrm dx=(-n)^{k+1}k!\zeta(k+2)?$ Proposed:
A simple closed form

$$\int_{0}^{1}{\ln^k(x)\over x}\ln\left(1-\sqrt[n]{x}\right)\mathrm dx=(-n)^{k+1}k!\zeta(k+2)\tag1$$

Where $n,k=1,2,3,\cdots$
My try:
$u=\sqrt[n]{x}\implies {nx\over u}du=dx$, $x=u^n$, then $(1)$ becomes
$$n^{k+1}\int_{0}^{1}{\ln^k(u)\over u}\ln(1-u)\mathrm du\tag2$$
$$-n^{k+1}\sum_{v\ge1}{1\over v}\int_{0}^{1}u^v\ln^k(u)\mathrm du\tag3$$
$$-n^{k+1}\sum_{v\ge1}{1\over v}\cdot{(-1)^kk!\over (v+1)^{k+1}}\tag4$$
$$(-n)^{k+1}k!\sum_{v\ge1}{1\over v(v+1)^{k+1}}\tag5$$
How may we prove $(1)?$
 A: With the change of variable $x=z^n$ you are left with something that clearly depends on the derivatives of a Beta function:
$$ \int_{0}^{1}\log(1-u)\log^k(u)\frac{du}{u} = -\lim_{\substack{\alpha\to 0^+\\ \beta\to 0^+}}\frac{d^k}{d\alpha^k}\frac{d}{d\beta}\int_{0}^{1}u^{\alpha-1}(1-u)^{\beta}\,du$$
and the RHS is just
$$ \lim_{\alpha\to 0^+}\frac{d^k}{d\alpha^k}\left(\frac{H_\alpha}{\alpha}\right) =(-1)^k k!\,\zeta(k+2)$$
by a straightforward computation through the Digamma function and its derivatives.
As an alternative,
$$\begin{eqnarray*} \sum_{n\geq 1}\frac{1}{n(n+1)^{k+1}}&=&\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+1}\right)\frac{1}{(n+1)^k}\\&\stackrel{SBP}{=}&\sum_{n\geq 1}\left(1-\frac{1}{n+1}\right)\frac{1}{(n+1)^k}-\sum_{n\geq 1}\left(1-\frac{1}{n+1}\right)\frac{1}{(n+2)^k}\\&=&1-\zeta(k+1)+\sum_{n\geq 1}\frac{1}{n(n+1)^k} \end{eqnarray*}$$
allows to compute OP's $(5)$ by induction:
$$ \sum_{n\geq 1}\frac{1}{n(n+1)^k} = k-\zeta(2)-\zeta(3)-\ldots-\zeta(k).$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
\int_{0}^{1}{\ln^{k}\pars{x} \over x}\ln\pars{1 - \root[\large n]{x}}\,\dd x &
\,\,\,\stackrel{\root[\large n]{x}\ \mapsto\ x}{=}\,\,\,
n^{k + 1}\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\ln^{k}\pars{x}\,\dd x
\\[5mm] & =
-n^{k + 1}\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{k}\pars{x}\,\dd x
\label{1}\tag{1}
\end{align}

Note that

\begin{align}
&\left.\int_{0}^{1}\mrm{Li}_{s}'\pars{x}\ln^{\ell}\pars{x}\,\dd x\,
\right\vert_{\ \ell\ >\ 0} =
-\ell\int_{0}^{1}\mrm{Li}_{s + 1}'\pars{x}\ln^{\ell - 1}\pars{x}\,\dd x
\\[5mm] = &\
\ell\pars{\ell - 1}\int_{0}^{1}\mrm{Li}_{s + 2}'\pars{x}
\ln^{\ell - 2}\pars{x}\,\dd x = \cdots =
\pars{-1}^{\ell}\ell\pars{\ell - 1}\cdots 1
\int_{0}^{1}\mrm{Li}_{s + \ell}'\pars{x}\,\dd x
\\[5mm] = &\
\pars{-1}^{\ell}\,\ell!\,\,\mrm{Li}_{s + \ell}\pars{1} =
\bbx{\pars{-1}^{\ell}\,\ell!\,\zeta\pars{s + \ell}}
\end{align}

Expression \eqref{1} becomes

$$
\int_{0}^{1}{\ln^{k}\pars{x} \over x}\ln\pars{1 - \root[\large n]{x}}\,\dd x =
-n^{k + 1}\bracks{\pars{-1}^{k}\,k!\,\zeta\pars{2 + k}} =
\bbx{\pars{-n}^{k + 1}\,k!\,\zeta\pars{k + 2}}
$$
