Have I followed through on this proof by induction correctly? We are asked to prove that $1^3+2^3+\cdots+n^3 = (1+2+\cdots + n)^2$ by induction.
Basis: $n=1 \\ 1^3 = 1^2 \quad\checkmark \\ n=2 \\ 1^3+2^3 = (1+2)^2 = 9 \quad\checkmark
$
We rewrite both series using the following summation notation:
$$\sum_{i=1}^{n}i^3 = \left(\sum_{i=1}^{n}i\right)^2$$
Inductive Hypothesis: 
We assume that this equality holds for an arbitrary value $k$ such that:
$$\sum_{i=1}^{k}i^3 = \left(\sum_{i=1}^{k}i\right)^2$$
Inductive Step: 
We next want to show that the equality holds for $k+1$ by rewriting the sum as:
$$1^3+2^3+\cdots+k^3+(k+1)^3 = \sum_{i=1}^{k}i^3+(k+1)^3=\left(\sum_{i=1}^{k}i\right)^2+(k+1)^3=\left(\sum_{i=1}^{k+1}i\right)^2$$
$$\square$$

Your last step is not justified, unless you just want to say that the result is trivial and be a bada%# mathematician. If not, then we need to recall that $$\sum_{i=1}^{k}i=\frac{k(k+1)}{2} $$
Continuing from the last line, $$\begin{align}\bigg(\sum_{i=0}^{k} i \bigg)^2 +(k+1)^3&= \frac{k^2(k+1)^2}{4} +(k+1)^3\\&=\frac{(k+1)^2}{4}\bigg[k^2+4(k+1)\bigg]\\&=\frac{(k+1)^2}{4}\bigg[ k^2+4k+4\bigg]\\&=\frac{(k+1)^2}{4}(k+1+1)^2\\&=\bigg[\frac{(k+1)(k+1+1)}{2}\bigg]^2\\&=\bigg(\sum_{i=0}^{k+1}i\bigg)^2\end{align}$$
