How to order ordinal numbers It is well-known that for $2$ ordinals $a$ and $b$ we have one of 
$$a \in b, a=b \text{ or } b\in a$$
which is equivalent to $a$ or $b$ being subset of the other.
My issue is: when I have some particular ordinals, are there any general tips and/or tricks to order them?
Suppose I have the ordinals $a=\omega^2+\omega$ and $b=\omega+\omega^2$. What should I do so that I order them? (This is just an example I made up, might not be too enlightening)
We have $b=\omega+\omega^2=\omega(1+\omega)=\omega^2$, while $a=\bigcup_{c<a}\omega^2+c$ so if I have to guess, I'd say that $a>b$. However, I'm feeling quite uncomfortable in doing those types of questions, so if anyone could provide some insight, I'd be very grateful!
Thanks in advance
 A: Cantor normal form is indeed the right way to think about this. Let me try to give an "organic" explanation of why.
First, let's think about the indecomposable ordinals. An ordinal $\alpha$ is indecomposable if whenever we write $$\alpha=\beta+\gamma$$ with $\gamma\not=0$ (equivalently, with $\beta\not=\alpha$) we in fact have $\alpha=\gamma$. Some examples:


*

*$\omega$ is indecomposable: this is because [Finite]+[Finite]=[Finite], so [Finite]+[?]=[Infinite] implies [?] is infinite.

*$\omega+\omega$ is not indecomposable: this is because I've split it into two pieces, the second piece is $\omega$, and $\omega<\omega+\omega$. So not every limit ordinal is indecomposable.

*$\omega^2$ is indecomposable again - this is a good exercise.
It turns out that the indecomposable ordinals are exactly the powers of $\omega$. Now the following facts are crucial:


*

*If $\beta$ is indecomposable and $\alpha<\beta$, then $\alpha+\beta=\beta$. Basically, indecomposable ordinals "absorb" smaller ordinals. Note that indecomposability is necessary: $\omega<\omega+\omega$, but $\omega+(\omega+\omega)\not=\omega+\omega$. By (usual) induction, $\alpha\cdot n+\beta=\beta$ as well for finite $n$.

*Any ordinal $\alpha$ can be written uniquely as a nonincreasing sum of indecomposable ordinals.  This is the Cantor normal form of that ordinal.
These facts are proved by transfinite induction. Once we have them in hand, we can use them to add two ordinals: first convert the two ordinals into Cantor normal form using the second bullet point, then "smoosh" them together as appropriate using the first bullet point.
For example, to add $\omega+\omega^3\cdot 17+\omega\cdot 3$ and $\omega^2+1$ we would say:
$$(\color{red}{\omega+\omega^3\cdot 17}+\omega\cdot 3)+(\omega^2+1)=(\omega^3\cdot 17+\omega\cdot 3)+(\omega^2+1)=\omega^3\cdot 17+\color{red}{\omega\cdot3+\omega^2}+1=\omega^3\cdot 17+\omega^2+1.$$ (I've highlighted in red the portions which get "absorbed.") Note that all the transfinite induction/recursion stuff has been relegated to proving those crucial facts mentioned above; once we have those, manipulating ordinals is more or less automatic.
