Prove $\alpha + \beta = \beta + \alpha \Rightarrow \alpha = \gamma n, \beta = \gamma m$ for some $m, n \in \mathbb{N}$. Here, $\alpha$ and $\beta$ are some ordinals for which addition turns out to be commutative, and $\gamma$ is some ordinal division by which yields a natural quotient and zero remainder for both $\alpha$ and $\beta$.
I tried proving this by assuming this is not the case (and, for every $\gamma$, dividing $\alpha$ and $\beta$ by $\gamma$ produces non-zero remainder if the quotient is a natural), but I couldn't come to any contradiction.
Another approach is to notice that if $\alpha + \beta = \beta + \alpha$, then for any sets $A, B$ representing respectively $\alpha, \beta$ with $L_A, L_B$ being the corresponding (well-ordered) sets of limit elements, $L_A + L_B = L_B + L_A$ must hold (since $A = \mathbb{N} \times L_A$ modulo some countable set of elements greater than the last limit element, and so is $B$). This is similar to the original claim, and one would hope that the ordinal of, say, $L_A$ is smaller than $\alpha$, so this could be proved inductively, but as far as I know this is not generally the case: the equality might also hold.
So what's the nicest way to prove the claim?
 A: Cantor normal form works.
Assume that we have following Cantor normal forms of $\alpha$ and $\beta$ respectively:
$$\alpha = \omega^{\xi_0}\cdot a_0+\cdots + \omega^{\xi_n}\cdot a_n\quad\text{ and } \beta = \omega^{\xi_0}\cdot b_0+\cdots + \omega^{\xi_n}\cdot b_n$$
(Of course, there is no need that their Cantor normal form have same length and same set of exponents. However such problem can be easily overcome by adding additional $\omega^\eta$-terms with coefficient zero.)
Now consider the following lemma: if $\xi<\eta$ and $k<\omega$ then $\omega^\xi\cdot k + \omega^\eta = \omega^\eta$. It follow from the existence of $\mu\ge 1$ such that $\eta = \xi + \mu$.
Assume that $\alpha+\beta = \beta+\alpha$ as given in our problem. We will divide cases: if $a_0=0$, then $\alpha+\beta = \beta < \beta + \alpha$ when $\alpha>0$. (The case when $\alpha=0$ is trivial.) Hence $\alpha+\beta\neq \beta+\alpha$. The case $b_0=0$ can be handled similarly.
Now assume that $a_0,b_0\neq 0$. Then the previous lemma leads following equation:
$$\omega^{\xi_0}\cdot (a_0+b_0) + \omega^{\xi_1}\cdot b_1 + \cdots \omega^{\xi_n}\cdot b_n = \omega^{\xi_0}\cdot (a_0+b_0) + \omega^{\xi_1}\cdot a_1 + \cdots \omega^{\xi_n}\cdot a_n $$
By left-cancellation law, we have
$$\omega^{\xi_1}\cdot b_1 + \cdots \omega^{\xi_n}\cdot b_n =  \omega^{\xi_1}\cdot a_1 + \cdots \omega^{\xi_n}\cdot a_n $$
Therefore $a_1=b_1$, $\cdots$, $a_n=b_n$ by the uniqueness of the Cantor normal form. Now take
$$\gamma = \omega^{\xi_0} + \omega^{\xi_1}\cdots a_1+\cdots + \omega^{\xi_n}\cdot a_n$$
then you can check that $\alpha = \gamma\cdot a_0$ and $\beta = \gamma\cdot b_0$.
