Prove $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$ using Induction Prove $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$ using Induction
My proof so far:
Let $P(n)$ be $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$

Base Case
$P(1):$
LHS = $1^3 = 1$
RHS = $(1)^2 = 1$
Since LHS = RHS, therefore base case holds

Induction Hypothesis
Let $n \in \mathbb{N}$  be arbitrary
Assume $P(n)$ holds

Induction Step
Prove $P(n+1)$ holds:
$$
\begin{align}
& 1^3 + 2^3 + \cdots + \;n^3 + (n+1)^3 \\
= {} & (1 + 2 + \cdots + \; n)^2 + (n+1)^3 \text{ (by Induction Hypothesis)} \\
= {} & (1 + 2 + \cdots + \; n)^2 + (n^3 + 3n^2 + 3n + 1)
\end{align}
$$

This is where I get stuck. I don't know how to prove that my last step is equivalent to:
$$(1 + 2 + \cdots + \;n + (n+1))^2$$
 A: Hint: by the binomial theorem,
$(1+2+\cdots +n +(n+1))^2=(1+2+\cdots +n)^2+(n+1)^2+2(n+1)(1+2+ \cdots n)$.
Now, you can show by induction (or by Gauss' trick) that $1+2+\cdots +n= \dfrac{n(n+1)}{2}$, and hus reach the answer.
A: So basically you want that last expression to turn out to be $\big(1+2+\ldots+n+(n+1)\big)^2$, so you want $(n+1)^3$ to be equal to the difference
$$\big(1+2+\ldots+n+(n+1)\big)^2-(1+2+\ldots+n)^2\;.$$
That’s a difference of two squares, so you can factor it as
$$(n+1)\Big(2(1+2+\ldots+n)+(n+1)\Big)\;.\tag{1}$$
To show that $(1)$ is just a fancy way of writing $(n+1)^3$, you need to show that
$$2(1+2+\ldots+n)+(n+1)=(n+1)^2\;.$$
Do you know a simpler expression for $1+2+\ldots+n$?
Work forward from this...
Okay so going forward.
Basically, Induction works like this, I'll use your question as an example:
Consider the case when $n = 1$. If so, then we will have $1^3 = 1^2$. 
Now suppose that $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for some $n \in \mathbb N$. 
Recall first that $\displaystyle (1 + 2 + 3 + \cdots + n) = \frac{n(n+1)}{2}$ so we know $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 = \bigg(\frac{n(n+1)}{2}\bigg)^2$. 
Now consider $\displaystyle 1^3 + 2^3 + 3^3 + \cdots + n^3 + (n + 1)^3 =  \bigg(\frac{n(n+1)}{2}\bigg)^2 + (n+1)^3 = \frac{n^2 (n+1)^2 + 4(n+1)^3}{4} = \bigg( \frac{(n+1)(n+2)}{2} \bigg)^2$. 
Hence, the statement holds for the $n + 1$ case. Thus by the  mathematical induction $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for each $n \in \mathbb N$.
$\mathbb Q.\mathbb E.\mathbb D$
