Proving $1^3+ \cdots + n^3 = (1 + \cdots + n)^2$ by induction

Question

I'm currently working on problems in Spivak's Calculus and just wanted to make sure that I have the right idea here. In Chapter 2, Problem 1, we are told to prove the following formula:

$$1^3+ \cdots + n^3 = (1 + \cdots + n)^2$$

Here is what I have:

This formula is true for $n = 1.$ Suppose it is true for $k$. Then

$$1^3+ \cdots + k^3 = (1 + \cdots + k)^2 \rightarrow 1^3+ \cdots + k^3 = \biggl(\frac {k(k+1)}{2}\biggl)^2.$$

So $$\begin{align}1^3+ \cdots + k^3 + (k+1)^3 & = \biggl(\frac {k(k+1)}{2}\biggl)^2+\,(k+1)^3 \\ & = \biggl(\frac{k^4+2k^3+k^2}{4} \biggl) + \,(k^3+3k^2+3k+1) \\ & = \frac{k^4+2k^3+k^2+4k^3+12k^2+12k+4}{4} \\ & = \frac{k^4+6k^3+13k^2+12k+4}{4} \\ & =\,\frac{(k+1)(k+2)(k+1)(k+2)}{4} \\ & =\,\biggl(\frac{(k+1)(k+2)}{2} \biggl)^2.\end{align}$$ $\therefore$ the formula holds true for $k+1$.

$$\tag*{$\blacksquare$}$$
Is there a better way to present this?

Answer 1

The base case and assumption are fine.

However, you can simplify your work for the inductive step.

You have that

$$1^3+2^3+\dots + k^3+ (k+1)^3 = \left(\dfrac{k(k+1)}{2}\right)^2+(k+1)^3.$$

Factoring $k+1$ out right away simplifies things. $$\begin{align}1^3+2^3+\dots + k^3+ (k+1)^3 = (k+1)^2\left(\dfrac{k^2+4k+4}{4}\right)\\ =\left(\dfrac{(k+1)(k+2)}{2}\right)^2\end{align}$$ and so by induction the statement holds for all $n\in\mathbb{N}$.