Proof of Equation by Well Ordering Principle I have an assignment question

Prove by either the Well Ordering Principle or induction that for all nonnegative integers $n$: $$\sum_{k=0}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2.$$

I am able to solve this question using basic Induction, but not able to figure out how to do it by using Well Ordering Principle.
Any Solutions or hints would be very helpful.
Thanks in Advance
 A: Suppose that the statement is not true. 
Then the set $A=\left\{ n\in\mathbb{N}\mid\sum_{i=0}^{n}i^{3}\neq\left(\frac{n\left(n+1\right)}{2}\right)^{2}\right\} $
is not empty. 
Since $\mathbb{N}$ is well-ordered set $A$ has a minimal
element $m$. 
That means that $\sum_{i=0}^{n}i^{3}=\left(\frac{n\left(n+1\right)}{2}\right)^{2}$
is true for $n<m$ and is not true for $n=m$. 
From this you can deduce
a contradiction.
(Start with $\sum_{i=0}^{m-1}i^{3}=\left(\frac{\left(m-1\right)m}{2}\right)^{2}$
and prove on base of that $\sum_{i=0}^{m}i^{3}=\sum_{i=0}^{m-1}i^{3}+m^{3}=\left(\frac{m\left(m+1\right)}{2}\right)^{2}$) 
The conclusion is then that $A=\emptyset$ wich is
exactly the statement to be proven.
A: Suppose that there are counterexamples. Take a minimal counterexample $n_0$...
A: For $n=1$ the left side is equal to $1$ and the right side is also equal to $1$ so the equality holds true.
Suppose that the equality $(1)$ holds true for some $n\geq1$ then for $n+1$ we obtain $$\sum_{i=1}^{n+1} i^3  =\sum_{i=1}^{n} i^3 +(n+1)^3 = \frac{n^2 (n+1)^2 }{4} +(n+1)^3 =(n+1)^2\left(\frac{1}{4} n^2 +n+1\right) =(n+1)^2 \left(\frac{1}{2} n +1\right)^2 =\left(\frac{(n+1)(n+2)}{2}\right)^2 .\fbox{}$$ 
