Concise induction step for proving $\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$ I recently got a book on number theory and am working through some of the basic proofs. I was able to prove that $$\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$$ with the help of the identity $$\sum_{i=1}^ni = \frac{n(n+1)}{2}$$ My proof of the induction step goes as follows (supposing equality holds for all $k \in \{1,2,\dots n \})$: $$\sum_{i=1}^{n+1} i^3 = \sum_{i=1}^{n} i^3+(n+1)^3 \\ =  \left( \sum_{i=1}^ni\right)^2+(n+1)^3 \\ =  \left( \frac{n(n+1)}{2}\right)^2 + (n+1)^3 \\ =  \frac{n^2(n+1)^2}{4}+\frac{4(n+1)^3}{4} \\ = \frac{n^4+2n^3+n^2}{4}+\frac{4n^3+12n^2+12n+4}{4} \\ = \frac{n^4+6n^3+13n^2+12n+4}{4} \\ = \frac{(n^2+3n+2)^2}{4} \\ = \frac{[(n+1)(n+2)]^2}{4} \\ = \left(\frac{(n+1)(n+2)}{2}\right)^2 \\ =  \left(\sum_{i=1}^{n+1}i\right)^2$$ I was a little disappointed in my proof because the algebra got really hairy. It took me a long while to see that I could twice unfoil the polynomial $n^4+6n^3+13n^2+12n+4$ and all in all the solution seems pretty inelegant. Does anyone have a smoother way to prove the induction step or bypass the algebra? I feel like their must be some concise way to get the same result.

We could also start the induction step from the RHS. This way we need only to cope with a square of a binom and not with a cube of a binom.

Induction step $n \rightarrow n+1$:
\begin{align*}
\left(\sum_{i=1}^{n+1}i\right)^2&=\left(\sum_{i=1}^{n}i+(n+1)\right)^2\\
&=\left(\sum_{i=1}^{n}i\right)^2 + 2(n+1)\sum_{i=1}^{n}i+(n+1)^2\\
&=\sum_{i=1}^{n}i^3 + 2(n+1)\frac{n(n+1)}{2}+(n+1)^2\tag{1}\\
&=\sum_{i=1}^{n}i^3 + (n+1)^3\\
&=\sum_{i=1}^{n+1}i^3\\
\end{align*}
In (1) we do the induction step and use the formula $\sum_{i=1}^ni=\frac{n(n+1)}{2}$

