Prove that $\left(\sum_{k=1}^{n}k\right)^2=\sum_{k=1}^{n}k^3$ holds true for $n ≥ 1$ I've been trying to figure out this proof for way too long now, I'm just not sure where to begin for the inductive step. Any guidance would be greatly appreciated.

Suppose it holds for some $n \geq 1$.
Claim: It holds for $n+1$, too.
Proof: We have
\begin{align*}
\left(1+ 2 + \ldots + n + (n+1) \right)^2 & = \left(1 + 2 + \ldots + n \right)^2 + 2 \left(1 + 2 + \ldots + n \right) (n+1) + (n+1)^2 \\
& = (1^3 + 2^3 + \ldots + n^3) + 2\frac{n(n+1)}{2} (n+1) + (n+1)^2 \\
& = (1^3 + 2^3 + \ldots + n^3) + n(n+1) ^2 + (n+1)^2 \\
& = 1^3 + 2^3 + \ldots + n^3 + (n+1)^3
\end{align*}
(I have made use of Gauss's addition forumla)
The assertion holds trivially for $n = 1$, it follows from induction that it holds for all $n \geq 1$.
