You might consider this to be a bit cheating, but here is a proof using only induction and one additional inference.
Lemma. If $A\subseteq\Bbb N$ has a maximal element, then $A$ has a minimal element.
Proof. First note that $A$ has a maximal element, so it is non-empty. Now we prove by induction on $n$: If $\max A=n$ then $A$ has a minimal element.
- If $\max A=0$ then $A=\{0\}$ and $0$ is also $\min A$.
- Assume that for $n$ the claim holds. If $\max A=n+1$ then either $A'=A\setminus\{n+1\}$ is empty, in which case $A=\{n+1\}$, so it is also the minimal element; else $\max A'=n$ and by the induction hypothesis it has a minimal element, $k$ and $k\leq n<n+1$, so $k=\min A$.
Therefore if $A$ has a maximal element, it has a minimal element. $\square$
Theorem. The principle of induction implies that every non-empty set of natural numbers has a minimal element.
Proof. If $A$ is a non-empty set, pick some $n\in A$, then $A'=\{a\in A\mid a\leq n\}$ has a maximal element $n$. We proved by induction that $A'$ has a minimum element $k$. If $m\in A$ then either $m\leq n$ in which case $m\in A'$ so $k\leq m$ or $m>n$ and in which case $k<m$. In either case $k=\min A$. $\square$
The Well-Ordering Principle states that any non-empty set can be well-ordered. This does not follow from the standard ZF axioms, and it is required to be assumed as its own axiom (conventionally, the Axiom of Choice is used instead, but assuming the rest of ZF, they are equivalent).
However, having well-ordered sets does not require the Well-Ordering Principle. Many sets can be proven to be well-orderable without any Well-Ordering Principle. This includes $\Bbb N$, as defined by your lecturer (assuming he has some sensible formal definition of $0$ and $+1$, and that he has shown that $\Bbb N$ is actually a set).
Similarily, proving that you can induct over $\Bbb N$ does not require the Well-Ordering Principle. However, proving that you can induct over any set does.
New answer: (After realizing I mixed up the Principle and the Theorem.)
You're mixing up the Well-Ordering Principle with the Well-Ordering Theorem. This is an easy thing to do (as witnessed by everything preceeding this paragraph). The Well-Ordering Principle is a theorem in ZF. It does not require any additional axioms, and it states that the standard ordering of $\Bbb N$ is a well-order.
The Well-Ordering Theorem is a statement that is independent of ZF, and states that all non-empty sets can be fitted with a well-order. This is what I described in my previous answering attempt.
Best Answer
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.