Multiple part problem concerning the proof that $\sum_{k=1}^n k^3=\left(\frac{n(n+1)}{2}\right)^2$ by induction 
So I'm having trouble with $c,d$ and $e$.
For $c$ so far I have:
Inductive Hypothesis: $(\frac{n(n+1)}{2})^2 = (\frac{(k+1)(k+2)}{2})^2$ is that correct?
 A: No. The inductive hypothesis is $\sum\limits_{k=1}^n k^3 = {({{\frac{{n(n+1)}}{2}}})}^2 $, which we assume is true.
From there:
$$
\sum\limits_{k=1}^n k^3 +(n+1) = {({\frac{n(n+1)}{2}})}^2+(n+1)
$$
Edit - the above has a typo, it should be:
$$
\sum\limits_{k=1}^n k^3 +(n+1)^3 = \left ({\frac{n(n+1)}{2}}\right )^2+(n+1)^3
$$
Try to see if you can go on from here.
A: For $n\geq 1$, let $P(n)$ denote the following statement:
$$
P(n) : \sum_{k=1}^n k^3=\left(\frac{n(n+1)}{2}\right)^2.
$$
(a): $P(1)$ is the statement that $\sum_{k=1}^1 k^3 = \left(\frac{1(1+1)}{2}\right)^2$. 
(b): $P(1)$ is true because $\sum_{k=1}^1 k^3 = 1 = \left(\frac{1(1+1)}{2}\right)^2$.
(c): Fix some $\ell\geq 1$ and assume that $P(\ell)$ is true where
$$
P(\ell) : \sum_{k=1}^\ell k^3=\left(\frac{\ell(\ell+1)}{2}\right)^2.
$$
That statement $P(\ell)$ is the inductive hypothesis.
(d): To be shown is that $P(\ell+1)$ follows where
$$
P(\ell+1) : \sum_{k=1}^{\ell+1} k^3=\left(\frac{(\ell+1)(\ell+2)}{2}\right)^2.
$$
Starting with the left-hand side of $P(\ell+1)$,
\begin{align}
\sum_{k=1}^{\ell+1}k^3 &= \color{red}{\sum_{k=1}^\ell k^3}+(\ell+1)^3\tag{by defn. of $\Sigma$}\\[1em]
&= \color{red}{\left(\frac{\ell(\ell+1)}{2}\right)^2}+(\ell+1)^3\tag{by $P(\ell)$, the ind. hyp}\\[1em]
&= \frac{(\ell+1)^2}{4}[\ell^2+4(\ell+1)]\tag{factor out $(\ell+1)^2/4$}\\[1em]
&= \frac{(\ell+1)^2}{4}[(\ell+2)(\ell+2)]\tag{factor quad. polynomial}\\[1em]
&= \frac{(\ell+1)^2(\ell+2)^2}{4}\tag{multiply and rearrange}\\[1em]
&= \left(\frac{(\ell+1)(\ell+2)}{2}\right)^2,
\end{align}
we end up at the right-hand side of $P(\ell+1)$. 
(e): Steps (a)-(d) show that the formula is true for all positive $n$ because we showed that $P(1)$ was true and that $P(\ell)\to P(\ell+1)$. This is what all we must show for mathematical induction to be a valid method of proof. We have shown this; thus, $P(n)$ must be true for all positive $n$. 
