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

It may be helpful to recognize that both the RHS and LHS represent the sum of the entries in a the multiplication tables. The LHS represents the summing of Ls (I'll outline those shortly), and the RHS, the summing of the sum of the rows [or columns])$$\begin{array}{lll}
\color{blue}\times&\color{blue}1&\color{blue}2\\
\color{blue}1&\color{green}1&\color{red}2\\
\color{blue}2&\color{red}2&\color{red}4\\
\end{array}$$
Lets begin by building our multiplication tables with a single entry, $1\times1=1=1^2=1^3$. Next, we add the $2$s, which is represented by the red L [$2+4+2 = 2(1+2+1)=2\cdot2^2=2^3$].
So the LHS (green 1 + red L) currently is $1^3+2^3$, and the RHS is $(1+2)+(2+4)=(1+2)+2(1+2)=(1+2)(1+2)=(1+2)^2$.
$$\begin{array}{llll}
\color{blue}\times&\color{blue}1&\color{blue}2&\color{blue}3\\
\color{blue}1&\color{green}1&\color{red}2&\color{maroon}3\\
\color{blue}2&\color{red}2&\color{red}4&\color{maroon}6\\
\color{blue}3&\color{maroon}3&\color{maroon}6&\color{maroon}9\\
\end{array}$$
Next, lets add the $3$s L. $3+6+9+6+3=3(1+2+3+2+1)=3\cdot3^2=3^3$. So now the LHS (green 1 + red L + maroon L) currently is $1^3+2^3+3^3$, and the RHS is $(1+2+3)+(2+4+6)+(3+6+9)=(1+2+3)+2(1+2+3)+3(1+2+3)=(1+2+3)(1+2+3)=(1+2+3)^2$.
By now, we should see a pattern emerging that will give us direction in proving the title statement.
Next we need to prove inductively that $\displaystyle\sum_{i=1}^n i = \frac{n(n+1)}{2}$, and use that relationship to show that $1+2+3+\dots+n+\dots+3+2+1 = \dfrac{n(n+1)}{2}+ \dfrac{(n-1)n}{2} = \dfrac{n((n+1)+(n-1))}{2}=\dfrac{2n^2}{2}=n^2$
Finally, it should be straight forward to show that:
$$\begin{array}{lll}
(\sum^n_{i=1}i+(n+1))^2 &=& (\sum^n_{i=1}i)^2 + 2\cdot(\sum^n_{i=1}i)(n+1)+(n+1)^2\\
 &=& \sum^n_{i=1}i^3 + (n+1)(\sum^n_{i=1}i + (n+1) + \sum^n_{i=1}i)\\
 &=& \sum^n_{i=1}i^3 + (n+1)(n+1)^2\\
 &=& \sum^n_{i=1}i^3 + (n+1)^3\\
 &=& \sum^{n+1}_{i=1}i^3\\
\end{array}$$
and, as was already pointed out previously, $$(\sum_{i=1}^1 i)^2 = \sum_{i=1}^1 i^3=1$$
