<s>[INST] Generalizing the sum of consecutive cubes $\sum_{k=1}^n k^3 = \Big(\sum_{k=1}^n k\Big)^2$ to other odd powers We have,
$$\sum_{k=1}^n k^3 = \Big(\sum_{k=1}^n k\Big)^2$$
$$2\sum_{k=1}^n k^5 = -\Big(\sum_{k=1}^n k\Big)^2+3\Big(\sum_{k=1}^n k^2\Big)^2$$
$$2\sum_{k=1}^n k^7 = \Big(\sum_{k=1}^n k\Big)^2-3\Big(\sum_{k=1}^n k^2\Big)^2+4\Big(\sum_{k=1}^n k^3\Big)^2$$
and so on (apparently). 
Is it true that the sum of consecutive odd $m$ powers, for $m>1$, can be expressed as sums of squares of sums* in a manner similar to the above? What is the general formula? 
*(Edited re Lord Soth's and anon's comment.)
 [/INST] This is not an answer but does not fit in a comment-box. 
Another way to describe the relation between the  sums and the sums of squares of various sums-of-like-powers makes use of the Eulerian-numbers, which converts the expressions for the sums-of-like-powers into polynomials.  The sums of like powers can be expressed as
$$\small \begin{array} {rclllll}  
s_0(n) = & 1 \cdot  \binom{n}{1}  \\ 
s_1(n) = & & 1 \cdot \binom{n+1}{2} \\ 
s_2(n) = & & 1 \cdot \binom{n+1}{3}& +1 \cdot \binom{n+2}{3} \\ 
s_3(n) = & & 1 \cdot \binom{n+1}{4}& +4 \cdot \binom{n+2}{4} & +1 \cdot \binom{n+3}{4} \\
s_4(n) = & & 1 \cdot \binom{n+1}{5}& +11 \cdot \binom{n+2}{5} & +11 \cdot \binom{n+3}{5} & +1 \cdot \binom{n+4}{5} \\ 
s_5(n) = & & 1 \cdot \binom{n+1}{6}& +26 \cdot \binom{n+2}{6}& +66 \cdot \binom{n+3}{6} & +26 \cdot \binom{n+4}{6} & +1 \cdot \binom{n+5}{6} \\ 
 \vdots &  &\vdots
 \end{array}
$$
This looks a bit more regular than the expressions by Bernoulli-polynomials and possibly the relations between the sums-expressions of odd orders $s_{2k+1}(n)$ can be written using these polynomials nicely and with  more ease than with the Bernoulli-polynomials, don't know.      
The first of your equality would then be expressed as
$$ \begin{array} {rcl} \sum_{k=1}^n k^3 &=&\left(\sum_{k=1}^n k \right)^2 \\  s_3(n) &=& s_1(n)^2 \\
 1 \cdot \binom{n+1}{4} +4 \cdot \binom{n+2}{4}  +1 \cdot \binom{n+3}{4} &=& \left( 1 \cdot \binom{n+1}{2}\right)^2 \\
 \end{array}$$ 
and which must then be expanded to reproduce the equality (but I've not the time at the moment to do this in the required length...)
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