Prove $\sum_{k=1}^nk^3=\sum_{n=1}^n\sum_{j=}^nk\cdot j$ 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
 A: By Induction
Per solution here, 
RHS=$\dfrac {n^2(n+1)^2}4$.
Proposition: 
$$\sum_{k=1}^n k^3=\sum_{k=1}^n\sum_{j=1}^n k\cdot j=\frac {n^2(n+1)^2}4$$
Induction:
Assume true for $n$. 
For $n+1$:
$$\sum_{k=1}^{n+1}k^3=\frac {n^2(n+1)^2}4+(n+1)^3=\frac {(n+1)^2}4\left(n^2+4(n+1)\right)=\frac {(n+1)^2(n+2)^2}4$$
i.e. also true for $n+1$. 
Clearly proposition is true for $n=1$. 
Hence, by induction, the proposition is true for all positive integer $n$. 

Direct Proof (Without using closed-form result)
This is a rather neat approach - a direct proof without having to work out the closed form. It is due to Misha Lavrov's answer  to a question I asked. 
$$\begin{align}
\sum_{k=1}^n k^3
&=\sum_{k=1}^n\sum_{j=1}^k k^2\\
&=\sum_{k=1}^n\sum_{j=1}^k k(j+(k-j))\\
&=\sum_{k=1}^n\left[\sum_{j=1}^k kj+\sum_{j=1}^k k(k-j)\right]\\
&=\sum_{k=1}^n\left[\sum_{j=1}^k kj+\sum_{i=0}^{k-1} ki\right]
&&\scriptsize(i=k-j)\\
&=\sum_{k=1}^n\left[\sum_{j=1}^k kj+\sum_{i=1}^{k-1} ki\right]
&&\scriptsize (ki=0\text{ when }i=0)\\
&=\sum_{k=1}^n\left[\sum_{j=1}^k kj+\sum_{j=1}^{k-1} kj\right]
&&\scriptsize\text{(using $j$ instead of $i$)}\\
&=\sum_{k=1}^n\sum_{j=1}^k kj+\sum_{j=1}^n\sum_{k=1}^{j-1}kj
&&\scriptsize\text{(swapping $j,k$ in second term)}\\
&=\sum_{k=1}^n\sum_{j=1}^k kj+\sum_{k=1}^n\sum_{j=k+1}^{n}kj
&&\scriptsize(1\le k<j\le n)\\
&=\sum_{k=1}^n\left(\sum_{j=1}^k kj+\sum_{j=k+1}^{n}kj\right)\\
&=\sum_{k=1}^n \sum_{j=1}^n kj\\
&=\sum_{k=1}^n k\sum_{j=1}^n j\\
&=\left(\sum_{k=1}^n k\right)^2\end{align}$$
A: 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*}
A: One can prove $ n^2+n=\sum_{k=1}^n2k$
aganin by induction.
$$\sum_{k=1}^{n+1}2k=n^2+n+2n+2$$
$$\sum_{k=1}^{n+1}2k=n^2+2n+1+n+1$$
$$\sum_{k=1}^{n+1}2k=\left(n+1\right)^2+n+1$$
This statement $B(n)$ is valid for all $n\in\mathbb{N}$.
Therefore, the first stament $A(n)$ must also be valid for all $n\in\mathbb{N}$.
A: For the value of $\sum k$ consider the following.
Starting with 
$$\sum_{k=1}^{n} x^{k} = \frac{x(1-x^{n})}{1-x}$$
and take a derivative to obtain
$$\sum_{k=1}^{n} k \, x^{k} = \frac{x}{(1-x)^{2}} \, (1 - (n+1) \, x^{n} + n \, x^{n+1})$$
and then use L'Hospital's rule to obtain
$$\sum_{k=1}^{n} k = \binom{n+1}{2}.$$
From here it can be shown that
$$\sum_{k=1}^{n} k^{3} = \frac{n^{2}(n+1)^{2}}{4} = \binom{n+1}{2}^{2} = \left(\sum_{k=1}^{n} k \right)^{2}.$$
