Induction proof for a summation: $sum_{i=1}^n i^3 = left[sum_{i=1}^n iright]^2$

Prove by induction: $\sum_{i=1}^n i^3 = \left[\sum_{i=1}^n i\right]^2$. Hint: Use $k(k+1)^2 = 2(k+1)\sum i$.

Basis: $n = 1$ $\sum_{i=1}^1 i^3 = \left[\sum_{i=1}^1 i\right]^2 \to 1^3 = 1^2 \to 1 = 1$.

Hypothesis: Assume true for all $n \le k$.

So far I have the following:

$$\sum_{i=1}^{k+1} i^3 = (k+1)^3 + \sum_{i=1}^k i^3$$

$$(k+1)^3 + \left[\sum_{i=1}^k i\right]^2$$

Answer

For $n=k+1$, $$\sum_{i=1}^{k+1}i^3 = \sum_{i=1}^{k}i^3+(k+1)^3=(\sum_{i=1}^{k}i)^2+(k+1)^3=(\sum_{i=1}^{k}i)^2+k(k+1)^2+(k+1)^2$$

Now using the Hint: $k(k+1)^2 = 2(k+1)\sum i$.

$$=(\sum_{i=1}^{k}i)^2+2(k+1)\sum_{i=1}^k i+(k+1)^2=(\sum_{i=1}^{k+1}i)^2$$