Sum of Sequence of Cubes/Proof by Recursion

Proof
From Closed Form for Triangular Numbers‎:
 * $(1): \quad \displaystyle \map A n := \sum_{i \mathop = 1}^n i = \frac{n \paren {n + 1} } 2$

From Sum of Sequence of Squares:
 * $(2): \quad \displaystyle \map B n := \sum_{i \mathop = 1}^n i^2 = \frac{n \paren {n + 1} \paren {2 n + 1} } 6$

Let $\displaystyle \map S n = \sum_{i \mathop = 1}^n i^3$.

Then: