Square of Triangular Number equals Sum of Sequence of Cubes/Proof 2

Proof
The proof proceeds by induction.

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

Basis for the Induction
$\map P 1$ is the case:

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 $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

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

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

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:
 * $\forall n \in \Z_{> 0}: \ds \sum_{i \mathop = 1}^n i^3 = {T_n}^2$