<s>[INST] 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.
 [/INST] 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 $n \geq 1$.
</s>