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

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 $(4)$.
