TITLE: iterated harmonic numbers vs Riemann zeta
QUESTION [13 upvotes]: Define the $m$-th iterated harmonic sums in the manner: $\bar{H}_0(n):=1$ and for 
$m\geq1$ by
$$\bar{H}_m(n):=\sum_{k=1}^n\frac{\bar{H}_{m-1}(k)}k.$$ 
For example, $\bar{H}_1(n)=\sum_{k=1}^n\frac1k$ are the familiar harmonic numbers. Euler proved that
$$\frac12\sum_{n\geq1}\frac{\bar{H}_1(n)}{n^2}=\zeta(3).$$
Hoping for a natural generalization, I ask:

Question 1. Is this true? If so, any proof?
  $$\frac1{m+1}\sum_{n\geq1}\frac{\bar{H}_m(n)}{n^2}=\zeta(m+2).$$

Of course, this works for $m=0$ as well: $\frac1{0+1}\sum_{n\geq1}\frac{\bar{H}_0(n)}{n^2}=\zeta(2)$.

Question 2. This might be of auxiliary interest. Any proof?
  $$\bar{H}_m(n)=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^m}\binom{n}k.$$

REPLY [3 votes]: Unless I messed up the notation, it seems that Note 5.3 of this paper

Luis A. Medina, Victor H. Moll, and Eric S. Rowland, Iterated primitives of logarithmic powers, International Journal of Number Theory 7 (2011) pp 623–634, doi:10.1142/S179304211100423X, arXiv:0911.1325 (pdf)

answers Question 2.
It seems quite likely that the OP is well aware of the above paper, given this recent paper:

Tewodros Amdeberhan, Christoph Koutschan, Victor H. Moll, Eric S. Rowland, The iterated integrals of $\ln(1 + x^n)$, International Journal of Number Theory 8 (2012) pp 71–94  doi:10.1142/S1793042112500042, arXiv:1012.3429 (.ps file).