Sum of Sequence of Cubes/Proof by Induction

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

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

Next we use induction on $n$ to show that:
 * $\ds \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$

The proof proceeds by induction.

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
 * $\ds \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$

Basis for the Induction
$\map P 1$ is the case:
 * $1^3 = \dfrac {1 \paren {1 + 1}^2} 4$

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\ds \sum_{i \mathop = 1}^k i^3 = \dfrac {k^2 \paren {k + 1}^2} 4$

from which it is to be shown that:
 * $\ds \sum_{i \mathop = 1}^{k + 1} i^3 = \dfrac {\paren {k + 1}^2 \paren {k + 2}^2} 4$

Induction Step
This is the induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

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