Sum of Sequence of Cubes/Proof using Multiplication Table

Proof
First, from Closed Form for Triangular Numbers:
 * $\displaystyle \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

So:
 * $\displaystyle \paren {\sum_{i \mathop = 1}^n i}^2 = \dfrac {n^2 \paren {n + 1}^2} 4$

This can be demonstrated using a simple Multiplication Table:


 * $\begin{array}{r|cccccccccc}

\paren {\sum_{i \mathop = 1}^n i}^2 & 1 & 2 & 3 & 4 & 5 & 6 & \cdots & N \\ \hline 1 & 1 & 2 & 3 & 4 & 5 & 6 & \cdots & N \\ 2 & 2 & 4 & 6 & 8 & 10 & 12 & \cdots & 2N  \\ 3 & 3 & 6 & 9 & 12 & 15 & 18 & \cdots & 3N \\ 4 & 4 & 8 & 12 & 16 & 20 & 24 & \cdots & 4N \\ 5 & 5 & 10 & 15 & 20 & 25 & 30 & \cdots & 5N \\ 6 & 6 & 12 & 18 & 24 & 30 & 36 & \cdots & 6N \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ N & N & 2N & 3N & 4N & 5N & 6N & \cdots & N^2 \\ \end{array}$

Consider the structure of each outermost edge of subsequent top left corner squares of the above of length $1$, $2$, $3$, $\ldots$.

We have:

Therefore:
 * $\forall n \in \Z_{>0}: \displaystyle \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$