Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction How can I prove that

$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$ 

for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks

Let P(n) be the given statement. You'll see why in the following step.
$$P(n):1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}$$
Step 1. Let $n = 1$.
Then $\mathrm{LHS} = 1^3 = 1$, $\mathrm{RHS} = \frac{1^2(1+1)^2}{4} = \frac{4}{4} = 1
$.
So LHS = RHS, and this means P(1) is true!
Step 2. Let $P(n)$ be true for $n = k$; that is,
$$1^3 + 2^3 + 3^3 + \cdots + k^3 = \frac{k^2(k+1)^2}{4}$$
We shall show that $P(k+1)$ is true too!
Add $(k+1)^3$, which is $(k+1)^{\mathrm th}$ term of the LHS to both sides of (1); then we get:
$$\begin{align*}
1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\
&= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\
&= \frac{(k+1)^2(k^2 + 4k + 4)}{4}\\
&= \frac{(k+1)^2(k+2)^2}{4}.\\
1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{(k+1)^2(k+2)^2}{4}
\end{align*}$$
I think this statement is the same as $P(n)$ with $n = k+1$.
