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
 A: We may notice that
$$ \sum_{n\geq 1}\frac{x^n}{n}=-\log(1-x),\qquad \sum_{n\geq 1} H_n x^n = -\frac{\log(1-x)}{1-x}\tag{1} $$
hence
$$ \sum_{n\geq 1}\frac{H_n}{n+1} x^{n}=\frac{\log^2(1-x)}{2x},\qquad \sum_{n\geq 1}\frac{H_{n+1}}{n+1} x^{n}=\frac{\log^2(1-x)+2\text{Li}_2(x)}{2x}-1 \tag{2}$$
and we may consider what the operator
$$ T_n=\sum_{k=0}^{n}(-1)^k \binom{n}{k}\frac{[x^k]}{k+1} \tag{3} $$
does to an analytic function in a neighbourhood of zero. This is strictly related with the binomial transform (and with Stirling numbers of the first kind giving the Taylor series of $\log(1-x)^k$), hence to solve the question it is enough to compute a closed form for 
$$ \sum_{k=0}^{n}(-1)^k \binom{n}{k}\frac{1}{(k+1)^3} = \frac{1}{2}\int_{0}^{1}\sum_{k=0}^{n}(-1)^k \binom{n}{k} x^k \log^2(x)\,dx = \frac{1}{2}\int_{0}^{1}(1-x)^n \log^2(x)\,dx$$
where the last integral is
$$ \frac{d^2}{d\alpha^2}\left.\int_{0}^{1}(1-x)^n x^{\alpha}\,dx\,\right|_{\alpha=0^+}=\frac{H_{n+1}^2+H_n^{(2)}}{2n+2}.\tag{4}$$
To finish the proof, it is enough to apply summation by parts to
$$ \sum_{j=0}^{k}\frac{H_{j+1}}{j+1} = H_{k+1}^{2}-\sum_{j=0}^{k-1}\frac{H_{j+1}}{j+1}.\tag{5} $$
Long story short: OP's identity come from applying the binomial transform to the series defining $\zeta(3)$ and exploiting my $(4)$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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$\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}
A: We seek to show that
$$\sum_{k=0}^n (-1)^k {n\choose k} \frac{1}{k+1}
\sum_{j=0}^k \frac{H_{j+1}}{j+1} = \frac{1}{(1+n)^3}.$$
This is
$$\sum_{k=0}^n (-1)^k {n+1\choose k+1} \frac{k+1}{n+1} \frac{1}{k+1}
\sum_{j=0}^k \frac{H_{j+1}}{j+1} = \frac{1}{(1+n)^3}$$
or
$$\sum_{k=0}^n (-1)^k {n+1\choose k+1} 
\sum_{j=0}^k \frac{H_{j+1}}{j+1} = \frac{1}{(1+n)^2}.$$
The LHS is
$$\sum_{j=0}^n \frac{H_{j+1}}{j+1} \sum_{k=j}^n (-1)^k {n+1\choose k+1}.$$
Writing
$${n+1\choose k+1} = {n+1\choose n-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+1}} (1+z)^{n+1}
\; dz$$
we get range control  (vanishes for $k\gt n$) so we  may write for the
inner sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{n+1}
\sum_{k\ge j} (-1)^k z^k
\; dz
\\ = (-1)^j \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-j+1}} (1+z)^{n+1}
\sum_{k\ge 0} (-1)^k z^k
\; dz
\\ = (-1)^j \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-j+1}} (1+z)^{n+1}
\frac{1}{1+z}
\; dz
\\ = (-1)^j \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-j+1}} (1+z)^{n}
\; dz
= (-1)^j {n\choose n-j} = (-1)^j {n\choose j}.$$
We thus have to show that
$$\sum_{j=0}^n \frac{H_{j+1}}{j+1} (-1)^j {n\choose j}
= \frac{1}{(1+n)^2}$$
or alternatively
$$\sum_{j=0}^n \frac{H_{j+1}}{j+1} (-1)^j 
\frac{j+1}{n+1} {n+1\choose j+1}
= \frac{1}{(1+n)^2}$$
which is
$$\sum_{j=0}^n H_{j+1} (-1)^j 
{n+1\choose j+1}
= \frac{1}{1+n}$$
The LHS is
$$\sum_{j=0}^n (-1)^j 
{n+1\choose j+1} \sum_{q=1}^{j+1} \frac{1}{q}
= \sum_{j=0}^n (-1)^j 
{n+1\choose j+1} \sum_{q=0}^{j} \frac{1}{q+1}
\\ = \sum_{q=0}^n \frac{1}{q+1}
\sum_{j=q}^n (-1)^j {n+1\choose j+1}.$$
We re-use the computation from before to get
$$\sum_{q=0}^n \frac{1}{q+1} (-1)^q {n\choose q}
= \sum_{q=0}^n \frac{1}{q+1} (-1)^q \frac{q+1}{n+1} {n+1\choose q+1}
\\ = \frac{1}{n+1} \sum_{q=0}^n (-1)^q {n+1\choose q+1}.$$
We have reduced the claim to
$$\sum_{q=0}^n (-1)^q {n+1\choose q+1} = 1$$
which holds by inspection or by writing 
$$- \sum_{q=1}^{n+1} (-1)^q {n+1\choose q}
= 1 - \sum_{q=0}^{n+1} (-1)^q {n+1\choose q} = 1 - (1-1)^{n+1} = 1.$$
