TITLE: Sum of powers of Harmonic Numbers
QUESTION [13 upvotes]: This is a natural extension of the question Sum of Squares of Harmonic Numbers. I became interested in this question while studying the problem A closed form of $\sum_{n=1}^\infty\left[ H_n^2-\left(\ln n+\gamma+\frac1{2n} \right)^2\right]$, especially the variation with a general integer power $q$, of which the present question is a part.
Let the sum in question be
$$s(q,n) = \sum _{k=1}^n H_k^{q} \tag{1}$$
where $q = 1, 2, 3, ...$ and $H_k = 1 + 1/2 + ... + 1/k$ is the harmonic number.
Problems

Derive a recursion relation for $s(q,n)$
Calculate $s(q,n)$ for $q=1 .. 4$
Discuss similarities with and possible deviation from the corresponding integrals $i(q,n)=\int_1^n ln(k)^q \, dk$ 
Discuss the possible meaning of the term "closed form" in the present context of finite sums.

I have given an answer to 1. and 2. up to q = 4, and 3. correspondingly.
Question 4. is perhaps the most interesting, and I have started to answer it.

REPLY [4 votes]: I'm coming back to this question after almost one year because I was able to find the complete answer. 
Recently, I became aware of the existence of the summation tool Sigma.m which is available as a Mathematica package (https://risc.jku.at/sw/sigma/).
This tool provides a framework for relating (harmonic) sums of different types, and I used it heuristically for the present task. Of course, also with this tool there is no "free dinner". I had to study several special cases, find a pattern and then make a good guess for the set of basic sums through wich a given sum can be expressed. The guess was then confirmed for a number of cases.
Main result
The complete answer to the question can be formulated as a follows
The finite sum of the q-th power of the harmonic number 
$$s_{q}(n) = \sum_{k=1}^n H_{k}^q,\;\; q = 1, 2, 3...\tag{1}$$
can be epressed in terms of a set of basic sums
$$t_{q} = \{H_n,\{H_{n}^{(r)},\; r=1..q\}, \{\sum_{k=1}^n \frac{H_{k}^{(r)}}{k}, \;{r=2..q}\},\\\{\sum_{k=1}^n \frac{H_{k}^r}{k^s},\;r+s=3..q-1,\; r,s,\ge 1,\; q\ge 4\}\}\tag{2}$$
Where, for given $q$, all basic sums whose parameters obey the ranges indicated must be taken into account.
This set is called "Tower" in the framework of Sigma. The main task was to find the correct form of the tower.
Proof
Up to now I have no formal proof. However, as the particular results for q=4 through q=10 are correct, I see it as a strong indication that the theorem is valid. The next task would be to find the expressions for the coefficients. 
Technical remarks
For Mathematica users here's the code for the Tower 
tow[q_] := Join[{SigmaHNumber[n]},
  Table[SigmaHNumber[m, n], {m, 2, q}],
  Table[SigmaSum[1/k SigmaHNumber[m, k], {k, 1, n}], {m, 2, q - 2}],
  Table[SigmaSum[1/k^m SigmaHNumber[k]^(q - 1 - m), {k, 1, n}], {m, 1,
     q - 2}],
  Flatten[
   Table[Table[
     SigmaSum[1/k^m SigmaHNumber[k]^(p + 1 - m), {k, 1, n}], {m, 1, 
      p}], {p, 2, q - 2}]]]

The outpunt $f$ of Sigma was transformed to Mathematica (and then to LaTEX) using the replacements
rep[f_] := 
 f /. Subscript[\[Iota], 1] -> k /. 
   Sigma`Summation`Objects`Private`MySum -> Sum /. 
  Sigma`Summation`Objects`Private`HNumber[a_, b_] -> 
   HarmonicNumber[b, a]

and xSum was subsequently replaced textually with Sum, finally LaTEX was taken from that expression. The use of  xSum was necessary to prevent Mathematica from (prematurely) calculating the expressions. 
The sums for $q=3..10$
Here comes the output of Sigma with the tower (2). Sorry for not having it "beautified" with respect to the order of the terms for now.
Notice that the cases $q=3$ und $q=4$ confirm the results obatained earlier "manually".
$$s_{3} = (n+1) \left(H_n\right){}^3-\frac{3}{2} (2 n+1) \left(H_n\right){}^2+3 (2 n+1) H_n+\frac{H_n^{(2)}}{2}-6 n$$
$$s_{4}=-2 \sum _{k=1}^n \frac{H_k^{(2)}}{k}+H_n \left(2 H_n^{(2)}-12 (2 n+1)\right)+(n+1) \left(H_n\right){}^4-2 (2 n+1) \left(H_n\right){}^3+6 (2 n+1) \left(H_n\right){}^2-2 H_n^{(2)}+H_n^{(3)}+24 n$$
$$s_{5}=-\frac{10}{3} \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k}+10 \sum _{k=1}^n \frac{H_k^{(2)}}{k}+\frac{5}{3} \sum _{k=1}^n \frac{H_k^{(3)}}{k}+H_n \left(-10 H_n^{(2)}-\frac{5 H_n^{(3)}}{3}+60 (2 n+1)\right)+(n+1) \left(H_n\right){}^5-\frac{5}{3} (3 n+2) \left(H_n\right){}^4+10 (2 n+1) \left(H_n\right){}^3-30 (2 n+1) \left(H_n\right){}^2+10 H_n^{(2)}-5 H_n^{(3)}-H_n^{(4)}-120 n$$
$$s_{6}=-5 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^2}-20 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k}+\frac{15}{2} \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k}-60 \sum _{k=1}^n \frac{H_k^{(2)}}{k}-10 \sum _{k=1}^n \frac{H_k^{(3)}}{k}-\frac{3}{2} \sum _{k=1}^n \frac{H_k^{(4)}}{k}+H_n \left(60 H_n^{(2)}+10 H_n^{(3)}+\frac{3 H_n^{(4)}}{2}-360 (2 n+1)\right)+(n+1) \left(H_n\right){}^6-\frac{3}{2} (4 n+3) \left(H_n\right){}^5+10 (3 n+2) \left(H_n\right){}^4-60 (2 n+1) \left(H_n\right){}^3+180 (2 n+1) \left(H_n\right){}^2-60 H_n^{(2)}+30 H_n^{(3)}+6 H_n^{(4)}+H_n^{(5)}+720 n$$
$$s_{7} = 7 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^3}+35 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^2}-14 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^2}+140 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k}-\frac{105}{2} \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k}+\frac{63}{5} \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k}+420 \sum _{k=1}^n \frac{H_k^{(2)}}{k}+70 \sum _{k=1}^n \frac{H_k^{(3)}}{k}+\frac{21}{2} \sum _{k=1}^n \frac{H_k^{(4)}}{k}+\frac{7}{5} \sum _{k=1}^n \frac{H_k^{(5)}}{k}+H_n \left(-420 H_n^{(2)}-70 H_n^{(3)}-\frac{21 H_n^{(4)}}{2}-\frac{7 H_n^{(5)}}{5}+2520 (2 n+1)\right)+(n+1) \left(H_n\right){}^7-\frac{7}{5} (5 n+4) \left(H_n\right){}^6+\frac{21}{2} (4 n+3) \left(H_n\right){}^5-70 (3 n+2) \left(H_n\right){}^4+420 (2 n+1) \left(H_n\right){}^3-1260 (2 n+1) \left(H_n\right){}^2+420 H_n^{(2)}-210 H_n^{(3)}-42 H_n^{(4)}-7 H_n^{(5)}-H_n^{(6)}-5040 n$$
$$s_{8} = -\frac{28}{3} \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^4}-56 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^3}+\frac{70}{3} \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^3}-280 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^2}+112 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^2}-28 \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k^2}-1120 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k}+420 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k}-\frac{504}{5} \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k}+\frac{56}{3} \sum _{k=1}^n \frac{\left(H_k\right){}^6}{k}-3360 \sum _{k=1}^n \frac{H_k^{(2)}}{k}-560 \sum _{k=1}^n \frac{H_k^{(3)}}{k}-84 \sum _{k=1}^n \frac{H_k^{(4)}}{k}-\frac{56}{5} \sum _{k=1}^n \frac{H_k^{(5)}}{k}-\frac{4}{3} \sum _{k=1}^n \frac{H_k^{(6)}}{k}+H_n \left(3360 H_n^{(2)}+560 H_n^{(3)}+84 H_n^{(4)}+\frac{56 H_n^{(5)}}{5}+\frac{4 H_n^{(6)}}{3}-20160 (2 n+1)\right)+(n+1) \left(H_n\right){}^8-\frac{4}{3} (6 n+5) \left(H_n\right){}^7+\frac{56}{5} (5 n+4) \left(H_n\right){}^6-84 (4 n+3) \left(H_n\right){}^5+560 (3 n+2) \left(H_n\right){}^4-3360 (2 n+1) \left(H_n\right){}^3+10080 (2 n+1) \left(H_n\right){}^2-3360 H_n^{(2)}+1680 H_n^{(3)}+336 H_n^{(4)}+56 H_n^{(5)}+8 H_n^{(6)}+H_n^{(7)}+40320 n$$
$$s_{9} = 12 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^5}+84 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^4}-36 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^4}+504 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^3}-210 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^3}+54 \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k^3}+2520 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^2}-1008 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^2}+252 \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k^2}-48 \sum _{k=1}^n \frac{\left(H_k\right){}^6}{k^2}+10080 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k}-3780 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k}+\frac{4536}{5} \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k}-168 \sum _{k=1}^n \frac{\left(H_k\right){}^6}{k}+\frac{180}{7} \sum _{k=1}^n \frac{\left(H_k\right){}^7}{k}+30240 \sum _{k=1}^n \frac{H_k^{(2)}}{k}+5040 \sum _{k=1}^n \frac{H_k^{(3)}}{k}+756 \sum _{k=1}^n \frac{H_k^{(4)}}{k}+\frac{504}{5} \sum _{k=1}^n \frac{H_k^{(5)}}{k}+12 \sum _{k=1}^n \frac{H_k^{(6)}}{k}+\frac{9}{7} \sum _{k=1}^n \frac{H_k^{(7)}}{k}+H_n \left(-30240 H_n^{(2)}-5040 H_n^{(3)}-756 H_n^{(4)}-\frac{504 H_n^{(5)}}{5}-12 H_n^{(6)}-\frac{9 H_n^{(7)}}{7}+181440 (2 n+1)\right)+(n+1) \left(H_n\right){}^9-\frac{9}{7} (7 n+6) \left(H_n\right){}^8+12 (6 n+5) \left(H_n\right){}^7-\frac{504}{5} (5 n+4) \left(H_n\right){}^6+756 (4 n+3) \left(H_n\right){}^5-5040 (3 n+2) \left(H_n\right){}^4+30240 (2 n+1) \left(H_n\right){}^3-90720 (2 n+1) \left(H_n\right){}^2+30240 H_n^{(2)}-15120 H_n^{(3)}-3024 H_n^{(4)}-504 H_n^{(5)}-72 H_n^{(6)}-9 H_n^{(7)}-H_n^{(8)}-362880 n$$
$$s_{10} =-15 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^6}-120 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^5}+\frac{105}{2} \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^5}-840 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^4}+360 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^4}-\frac{189}{2} \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k^4}-5040 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^3}+2100 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^3}-540 \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k^3}+105 \sum _{k=1}^n \frac{\left(H_k\right){}^6}{k^3}-25200 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^2}+10080 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^2}-2520 \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k^2}+480 \sum _{k=1}^n \frac{\left(H_k\right){}^6}{k^2}-75 \sum _{k=1}^n \frac{\left(H_k\right){}^7}{k^2}-100800 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k}+37800 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k}-9072 \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k}+1680 \sum _{k=1}^n \frac{\left(H_k\right){}^6}{k}-\frac{1800}{7} \sum _{k=1}^n \frac{\left(H_k\right){}^7}{k}+\frac{135}{4} \sum _{k=1}^n \frac{\left(H_k\right){}^8}{k}-302400 \sum _{k=1}^n \frac{H_k^{(2)}}{k}-50400 \sum _{k=1}^n \frac{H_k^{(3)}}{k}-7560 \sum _{k=1}^n \frac{H_k^{(4)}}{k}-1008 \sum _{k=1}^n \frac{H_k^{(5)}}{k}-120 \sum _{k=1}^n \frac{H_k^{(6)}}{k}-\frac{90}{7} \sum _{k=1}^n \frac{H_k^{(7)}}{k}-\frac{5}{4} \sum _{k=1}^n \frac{H_k^{(8)}}{k}+H_n \left(302400 H_n^{(2)}+50400 H_n^{(3)}+7560 H_n^{(4)}+1008 H_n^{(5)}+120 H_n^{(6)}+\frac{90 H_n^{(7)}}{7}+\frac{5 H_n^{(8)}}{4}-1814400 (2 n+1)\right)+(n+1) \left(H_n\right){}^{10}-\frac{5}{4} (8 n+7) \left(H_n\right){}^9+\frac{90}{7} (7 n+6) \left(H_n\right){}^8-120 (6 n+5) \left(H_n\right){}^7+1008 (5 n+4) \left(H_n\right){}^6-7560 (4 n+3) \left(H_n\right){}^5+50400 (3 n+2) \left(H_n\right){}^4-302400 (2 n+1) \left(H_n\right){}^3+907200 (2 n+1) \left(H_n\right){}^2-302400 H_n^{(2)}+151200 H_n^{(3)}+30240 H_n^{(4)}+5040 H_n^{(5)}+720 H_n^{(6)}+90 H_n^{(7)}+10 H_n^{(8)}+H_n^{(9)}+3628800 n $$
Discussion
First of all, the strict proof would be desirable. I invite others herewith to contribute.
There are several natural questions emerge from this result are, for instance: 
a) is the set $t_{q}$ minimal?     
b) are the sums in $t_{q}$ really basic or can the be further reduced?    
c) are there equiavalent sets to (2)? We have already confirmed this for $q=4$ in the first answer.    
d) can we find explicit expressions for the coefficents and hence construct the explicit closed expression for $s_{q}$? Note that the number of basic sums in $s_{q}$ is given by $\frac{1}{2} (q-2)(q-3)$.
Generalizations could include modified harmonic sums.