TITLE: Prove $\sum_{k=1}^nk^3=\sum_{n=1}^n\sum_{j=}^nk\cdot j$
QUESTION [6 upvotes]: I am trying to prove $$\sum_{k=1}^nk^3=\sum_{k=1}^n\sum_{j=1}^nk\cdot j$$ by using induction. Alternative approaches to this problem are possible as well.
My thoughts so far:
$\sum_{k=1}^{n+1}k^3=\sum_{k=1}^{n+1}\sum_{j=1}^{n+1}k\cdot j$
$\Rightarrow\sum_{k=1}^nk^3+\left(n+1\right)^3=\sum_{k=1}^{n+1}k\cdot\sum_{j=1}^{n+1}j=\left(\sum_{k=1}^nk+n+1\right)\cdot\left(\sum_{j=1}^nj+n+1\right)$
$\Rightarrow\left(n+1\right)^3=\left(n+1\right)^2+\left(n+1\right)\cdot\left(\sum_{k=1}^nk+\sum_{j=1}^nj\right) $
$\Rightarrow \left(n+1\right)^2=\left(n+1\right)^{ }+\sum_{k=1}^n2k$    
$\Rightarrow n^2+n=\sum_{k=1}^n2k$
This is where I am stuck. How can I continue from here? I read something about geometric progressions; do they apply to this example?
Any help is appreciated.
Philipp

REPLY [2 votes]: You're almost done. To conclude simply note that
\begin{align*}
2\sum_{k=1}^nk&=({\color{red}1}+{\color{blue}2}+\ldots+{\color{green}n})+({\color{red}n}+{\color{blue}{n-1}}+\ldots+{\color{green}1})
\\ &=({\color{red}1}+{\color{red}n})+({\color{blue}2}+{\color{blue}{n-1}})+\ldots + ({\color{green}n}+{\color{green}1})\\ &=n(n+1)\end{align*}