Does Fermat's last theorem hold in the ordinals? My question is whether there are no nontrivial solutions in the ordinals of the equations arising in Fermat's last theorem $$x^n+y^n=z^n$$
where $n\gt 2$, and where we use the  natural ordinal arithmetic, which is commutative. 
(Note: If we had used the usual ordinal arithmetic, there are some easy counterexamples, such as $1^3+\omega^3=\omega^3$.)
The question spins off of my answer to Saint Georg's recent question, The Theory of Transfinite Diophantine Equations.  
Feldmann Denis pointed out in the comments there that for very small ordinals (below $\omega^\omega$), the question reduces to the corresponding question in polynomials, where it has an affirmative answer. Can we extend this observation to work for all ordinals? 
 A: The review (by Erdos) of Sierpinski, Le dernier théorème de Fermat pour les nombres ordinaux, Fund. Math. 37 (1950) 201–205, MR0040372 (12,683c), says, "...there are arbitrarily large transfinite ordinals $\alpha,\beta,\gamma$, $\alpha\lt\beta\lt\gamma$ so that for all $n=1,2,\dots$, $\alpha^n+\beta^n=\gamma^n$. I haven't looked at the paper, so I don't know whether this is in natural or usual ordinal arithmetic. 
A: There are no nontrivial solutions. This follows from Wiles’s proof, and the following observation.
Proposition: If a set of Diophantine equations has a solution in (positive) ordinals using natural sum and product, then it has a solution in (positive) natural numbers.
Proof: Every ordinal can be uniquely written in Cantor normal form
$$\tag{$*$}\alpha=\sum_{i<k}\omega^{\alpha_i}n_i,$$
where $\alpha_i>\alpha_{i+1}$ and $0<n_i<\omega$ for all $i<k$. Define a function $f\colon\mathrm{Ord}\to\omega$ by $f(\alpha)=\sum_jn_j$. Notice that $f(\alpha)=0$ only if $\alpha=0$, and $f(n)=n$ for all $n<\omega$. The proposition follows from
Claim: $f$ is a homomorphism with respect to natural sum and product.
This in turn follows from expression of the operations in terms of Cantor normal form. For sum, let $\alpha$ and
$$\beta=\sum_i\omega^{\alpha_i}m_i$$
be as in $(*)$, except that we allow $n_i$ and $m_i$ to be zero. Then their natural sum is
$$\def\ns{\mathbin\#}\alpha\ns\beta=\sum_i\omega^{\alpha_i}(n_i+m_i),$$
hence
$$f(\alpha\ns\beta)=\sum_i(n_i+m_i)=f(\alpha)+f(\beta).$$
As for product, the expression in $(*)$ is valid even if we interpret the sum and product there as the natural operations, and natural product is associative and distributive over natural sum, hence it suffices to consider only the case of $\alpha=\omega^{\alpha_0}$ and $\beta=\omega^{\beta_0}$. However, then their natural product is $\gamma=\omega^{\alpha_0\ns\beta_0}$, and $f(\gamma)=1=f(\alpha)f(\beta)$.
