<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] To make this proof easier to follow, let's define the following:  $$\begin{align*} A_n &= \sum_{k=1}^n k, \\ B_n &= A_n^2 = \biggl(\sum_{k=1}^n k\biggr)^2, \\ C_n &= \sum_{k=1}^n k^3. \end{align*}$$  Then the claim to be proven is that $B_n = C_n$ for all positive integers $n$.  Here, we focus on the inductive step:  that is, we wish to establish that if there exists a positive integer $n$ such that $B_n = C_n$, this implies that $B_{n+1} = C_{n+1}$.  To this end, we first observe $A_{n+1} = (n+1) + A_n$:  hence $$\begin{align*} B_{n+1} &= A_{n+1}^2 = ((n+1) + A_n)^2 \\ &= (n+1)((n+1) + 2A_n) + A_n^2 \\ &= (n+1)(A_n + (n+1) + A_n) + B_n \\ &= (n+1)(A_{n+1} + A_n) + C_n. \end{align*}$$  But note $$A_{n+1} + A_n = (n+1) + \sum_{k=1}^n k + \sum_{k=n}^1 ((n+1)-k) = (n+1) + \sum_{k=1}^n (n+1) = (n+1)^2,$$ thus $$B_{n+1} = (n+1)^3 + C_n = C_{n+1},$$ which completes the inductive step.
</s>