Proving the identity $\sum_{k=1}^n {k^3} = \big(\sum_{k=1}^n k\big)^2$ without induction I recently proved that
$$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$$
using mathematical induction.  I'm interested if there's an intuitive explanation, or even a combinatorial interpretation of this property. I would also like to see any other proofs.

One way to show that
$$\sum_{i=1}^n i^3 = \bigg(\sum_{i=1}^n i \bigg)^2$$
is to add up the entries in the multiplication tables, but first we need to show that
$$1+2+3+\dots+n+\dots+3+2+1 = n^2$$
For this, see the image below (n=7)
$$7^2=\color{green}{1+2+3+4+5+6+7}\color{red}{+6+5+4+3+2+1}$$
Next, consider the standard multiplication table that we are all familiar with.The graphic shows the table up to the 9s.

We can add up the entries in any order that we wish.
One way would be to add up a series of Ls (the 6th L ($L_6$) is highlighted in yellow).
$$\begin{align}
L_6 &= 6+12+18+24+30+36+30+24+18+12+6\\
&=6(1+2+3+4+5+6+5+4+3+2+1)\\
&=6(6^2)\\
&=6^3
\end{align}$$
And the sum of all the entries in the table becomes
$$\sum_{i=1}^n L_i = \sum_{i=1}^n i^3$$
Alternatively, we could just add up each row. The 6th row ($R_6$) would be
$$\begin{align}
R_6 &= 6+12+18+24+30+36+42+48+54\\
&= 6(1+2+3+4+5+6+7+8+9)\\
&= 6\sum_{i=1}^9 i
\end{align}$$
And the sum of all the entries becomes
$$\sum_{i=1}^n R_i = \sum_{i=1}^n \bigg(i\sum_{j=1}^n j\bigg)=\bigg(\sum_{j=1}^n j\bigg)\bigg(\sum_{i=1}^n i\bigg)=\bigg(\sum_{i=1}^n i\bigg)^2$$
Thus we have
$$\sum_{i=1}^n i^3 = \sum_{i=1}^n L_i = \sum_{i=1}^n R_i=\bigg(\sum_{i=1}^n i\bigg)^2$$
