Showing that $\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}\sum_{j=0}^{k}{H_{j+1}\over j+1}={1\over (n+1)^3}$ Consider this double sums $(1)$

$$\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}\sum_{j=0}^{k}{H_{j+1}\over j+1}={1\over (n+1)^3}\tag1$$
  Where $H_n$ is the n-th harmonic

An attempt:
Rewrite $(1)$ as
$$\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}\left(H_1+{H_2\over 2}+{H_3\over 3}+\cdots+{H_{k+1}\over k+1}\right)\tag2$$
Recall $$\sum_{k=0}^{n}(-1)^k{n\choose k}{1\over k+1}={1\over n+1}\tag3$$
Not sure how to continue

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$\ds{\sum_{k = 0}^{n}\pars{-1}^{k}{n\choose k}{1 \over k + 1}
\sum_{j = 0}^{k}{H_{j + 1} \over j + 1} =
{1 \over \pars{n + 1}^{3}}:\ {\large ?}}$.  

With the Harmonic Number Generating Function $\ds{\,\mc{H}\pars{z} \equiv -\,{\ln\pars{1 - z} \over1 - z} = \sum_{n = 0}^{\infty}H_{n}z^{n}}$, lets consider the over-$\ds{j}$ sum: 
\begin{align}
\sum_{j = 0}^{k}{H_{j + 1} \over j + 1} & =
\sum_{j = 0}^{k}{\bracks{z^{\,j + 1}}\mc{H}\pars{z} \over j + 1} =
\bracks{z^{0}}\mc{H}\pars{z}\sum_{j = 0}^{k}{\pars{1/z}^{\,j + 1} \over j + 1} =
\bracks{z^{0}}\mc{H}\pars{z}\sum_{j = 0}^{k}\int_{0}^{1/z}x^{\,j}\,\dd x
\\[5mm] & =
\bracks{z^{0}}\mc{H}\pars{z}\int_{0}^{1/z}
{x^{k + 1} - 1 \over x - 1}\,\dd x
\end{align}

Then,
\begin{align}
&\sum_{k = 0}^{n}\pars{-1}^{k}{n\choose k}{1 \over k + 1}
\sum_{j = 0}^{k}{H_{j + 1} \over j + 1} =
-\sum_{k = 0}^{n}{n\choose k}{\pars{-1}^{k + 1} \over k + 1}
\braces{\bracks{z^{0}}\mc{H}\pars{z}\int_{0}^{1/z}
{x^{k + 1} - 1 \over x - 1}\,\dd x}
\\[5mm] = &\
\bracks{z^{0}}\mc{H}\pars{z}\int_{0}^{1/z}
\bracks{%
\sum_{k = 0}^{n}{n\choose k}{\pars{-x}^{k + 1} \over k + 1} -
\sum_{k = 0}^{n}{n\choose k}{\pars{-1}^{k + 1} \over k + 1}}{\dd x \over 1 - x}
\\[5mm] = &\
\bracks{z^{0}}\mc{H}\pars{z}\int_{0}^{1/z}
\bracks{\pars{1 - x}^{n + 1} \over n + 1}{\dd x \over 1 - x} =
{1 \over n + 1}\bracks{z^{0}}\mc{H}\pars{z}\int_{0}^{1/z}\pars{1 - x}^{n}\,\dd x
\\[5mm] = &\ 
{1 \over n + 1}\bracks{z^{0}}\mc{H}\pars{z}
{1 + \pars{-1}^n\pars{1/z - 1}^{n + 1} \over n + 1}
\\[5mm] = &\
{1 \over \pars{n + 1}^{2}}\braces{\vphantom{\Large A}%
\bracks{z^{0}}\mc{H}\pars{z} +
\pars{-1}^{n}\bracks{z^{n + 1}}\mc{H}\pars{z}\pars{1 - z}^{n + 1}}
\\[5mm] = &\
{\pars{-1}^{n + 1} \over \pars{n + 1}^{2}}
\bracks{z^{n + 1}}\pars{1 - z}^{n}\ln\pars{1 - z} =
{\pars{-1}^{n + 1} \over \pars{n + 1}^{2}}\bracks{z^{n + 1}}
\left.\partiald{\pars{1 - z}^{n + \mu}}{\mu}\right\vert_{\ \mu\ =\ 0}
\\[5mm] = &\
{\pars{-1}^{n + 1} \over \pars{n + 1}^{2}}
\left.\partiald{}{\mu}{n + \mu \choose n + 1}\pars{-1}^{n + 1}
\right\vert_{\ \mu\ =\ 0} = \bbx{\ds{1 \over \pars{n + 1}^{3}}}
\end{align}


Note that

\begin{align}
\partiald{}{\mu}{n + \mu \choose n + 1} & =
{n + \mu \choose n + 1}\pars{H_{n + \mu} - H_{\mu - 1}} =
{-\mu \choose n + 1}\pars{-1}^{n + 1}\pars{H_{n + \mu} - H_{\mu - 1}}
\\[5mm] & =
\pars{-1}^{n + 1}\,
{\Gamma\pars{1 - \mu} \over \pars{n + 1}!\,\Gamma\pars{-\mu - n}}
\bracks{H_{n + \mu} - H_{-\mu} + \pi\cot\pars{\pi\mu}}
\\[5mm] & =
\pars{-1}^{n + 1}\,
{\Gamma\pars{1 - \mu}\bracks{H_{n + \mu} - H_{-\mu} + \pi\cot\pars{\pi\mu}} \over
\pars{n + 1}!\pars{\pi/\braces{\Gamma\pars{n + 1 + \mu}
\sin\pars{\pi\bracks{n + 1 + \mu}}}}}
\\[5mm] & =
\underbrace{\Gamma\pars{1 - \mu}}_{\ds{\to\ 1\ \mrm{as}\ \mu\ \to\ 0}}\,\
\underbrace{{\Gamma\pars{n + 1 + \mu} \over \pars{n +1}!}}
_{\ds{\to\ {1 \over n + 1}\ \mrm{as}\ \mu\ \to\ 0}}\,\
\underbrace{\bracks{{\sin\pars{\pi\mu}\pars{H_{n + \mu} - H_{-\mu}} \over \pi} + \cos\pars{\pi\mu}}}_{\ds{\to\ 1\ \mrm{as}\ \mu\ \to\ 0}}
\end{align}
