$\alpha^{\beta}= \alpha.\beta$.
In order to do this, it is suggested to prove that for all $\alpha, \beta >2$, $\alpha^{\beta}= \alpha.\beta$ iff $\alpha^{\beta}= \beta$.
Here's my proof of this statement :
If $\alpha^{\beta}= \beta$ then $\alpha.\beta= \alpha.\alpha^{\beta}= \alpha^{1+\beta}$.
Notice that $1+\beta = \beta$ iff $\beta$ is infinite. If $\beta$ was finite ($<\omega$), then $\beta=n>2$ and we would have $\alpha^{1+n}= \alpha.\alpha^n\neq \alpha^{n}$ for all $\alpha>2$. Hence if $\beta$ is infinite and $\alpha^{\beta}= \beta$ then $\alpha.\beta= \alpha.\alpha^{\beta}= \alpha^{\beta}$.
Now, suppose that $\alpha^{\beta}= \alpha.\beta$.
Notice that again if $\beta$ was finite, then $2<\beta=n<\omega$ and we would have $\alpha^n=\alpha.n$ which is false for all $\alpha>2$. Hence $\beta$ is infinite. At this point, we know that $\beta$ is successor or $\beta$ is limit.
If $\beta$ is successor then there exists $\delta$ such that $\beta = \delta +1$. Hence $\alpha^{\beta}=\alpha^{\delta +1}= \alpha.{(\delta+1)}\ge \delta +1$ (order preservation on the left with $\alpha>1$).
If there is some $\alpha$ such that : $\alpha.(\delta+1)=\alpha$ we are done. Otherwise $\beta$ is limit. Let us say that $\beta= \sup_{\gamma<\beta}{\gamma}$.
Hence $\alpha^{\beta}=\sup_{\gamma<\beta}{\alpha^{\gamma}}=\sup_{\gamma<\beta}\gamma$ (induction hypothesis). Finally $\alpha^{\beta}=\beta$ and the statement is again proved.
I do not know if my proof is totally correct. Moreover how to understand "$\beta$ arbitrarily large" in the main statement ? Any hints or improvements for the proof of this property ?
Thanks in advance !