Ordinal exponentiation identity with natural sum of exponents This is related to a previous question on How to think about ordinal exponentiation?
One possible definition for the natural product $\alpha\otimes\beta$ of ordinals is based on Cantor Normal Forms and natural sum: if $\alpha=\omega^{\alpha_1}+\cdots+\omega^{\alpha_n}$ and $\beta=\omega^{\beta_1}+\cdots+\omega^{\beta_m}$ are CNFs, then
$\alpha\otimes\beta=\bigoplus_{i=1}^n\bigoplus_{j=1}^m\omega^{\alpha_i\oplus\beta_j}$.
With this definition, one automatically gets the following exponentiation law for natural products of exponentials: $\omega^\alpha\otimes\omega^\beta=\omega^{\alpha\oplus\beta}$.
My question: does it more generally hold that $\gamma^\alpha\otimes\gamma^\beta=\gamma^{\alpha\oplus\beta}$ for any $\gamma>0$? And if yes what do you recommend as a good reference?
I expected to find the answer on wikipedia page on ordinal arithmetic or on some other widely available source but did not manage.
 A: The answer is NO: it does not generally hold that $\gamma^\alpha\otimes\gamma^\beta=\gamma^{\alpha\oplus\beta}$.
For example, taking $\alpha=\beta=1$, we don't have $\gamma\otimes\gamma=\gamma^2$. Try it for $\gamma=\omega^2+\omega+1$. This gives
$$\begin{aligned}\gamma^2=\gamma\cdot\gamma&=(\omega^2+\omega+1)\cdot \omega^2 +
(\omega^2+\omega+1)\cdot\omega +
(\omega^2+\omega+1) \\&=
\omega^4+\omega^3+(\omega^2+\omega+1)\end{aligned}$$ while
$\gamma\otimes\gamma =\omega^4+\omega^3\cdot 2+\omega^2\cdot 3+\omega\cdot 2+1$.
One only has $\gamma^\alpha\otimes\gamma^\beta\geq\gamma^{\alpha\oplus\beta}$ in general.
PS: It seems that the equality holds (for any exponents $\alpha$ and $\beta$) when $\gamma$ is a principal ordinal (also called indecomposable ordinal, of the form $\omega^\delta$) and also when it is a finite ordinal. I previously thought that it also holds when $\gamma$ is a finite multiple of an indecomposable ordinal, which would be a common generalization of the two cases, but even this does not work: take e.g. $\gamma=\omega 2$ and $\alpha=\beta=\omega+1$.
