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

[All: This answer came here when I merged duplicates. Please take that into account when voting, JL]
You should prove before
$$
\sum_{i=0}^n i = \frac{n(n+1)}{2}
$$
Cases $n=0,1$ are trivial. Suppose it's true for $n-1$ then
\begin{align}
\sum_{i=0}^n i^3 &= n^3 + \sum_{i=0}^{n-1}i^3 \\
&=n^3 + \left(\sum_{i=0}^{n-1} i\right)^2 \\
&= n^3 + \frac{n^2(n-1)^2}{4}\\
& = \frac{4n^3 + n^4 + n^2 + -2n^3}{4}\\
&= \frac{(n+1)^2n^2}{4} \\
&= \left(\sum_{i=0}^{n} i\right)^2
\end{align}
