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 $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{i \mathop = 1}^n i^3 = {T_n}^2$

Basis for the Induction
$P \left({1}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.

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

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

Induction Step
This is the induction step:

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

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