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.
 A: An idea for your fourth line:
$$\frac{n^2(n+1)^2}{4}+\frac{4(n+1)^3}{4}=\frac{(n+1)^2}4\left(n^2+4n+4\right)=\frac{(n+1)^2}4(n+2)^2$$
A: This is indeed possible. Starting from your fourth line: $$\frac{n^2(n+1)^2}{4} + \frac{4(n+1)^3}{4} = \frac{n^2(n+1)^2 + 4(n+1)(n+1)^2}{4} = \frac{(n^2 + 4(n+1))(n+1)^2}{4}$$ $$=\frac{(n^2 + 4n + 4)(n+1)^2}{4} = \frac{(n+2)^2(n+1)^2}{4}. $$
A: You could have made it a little easier here:
$$
\frac{n^2(n+1)^2}{4}+\frac{4(n+1)^3}{4} = (n+1)^2 \left( \frac{n^2}{4}+\frac{4(n+1)}{4} \right) = \frac{(n+1)^2(n^2+4n+4)}{4} = \frac{((n+1)(n+2))^2}{4}.
$$
A: 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 $\sum_{i=1}^ni=\frac{n(n+1)}{2}$

A: Induction $n-1\to n$ makes it more compact.
$$\left(\sum_1^{n-1}i+n\right)^2=\left(\sum_1^{n-1}i\right)^2+2n\sum_1^{n-1}i+n^2=\left(\sum_1^{n-1}i\right)^3+n^2(n-1)+n^2=\left(\sum_1^{n-1}i\right)^3+n^3.$$
