Set Theory - Ordinal Exponents - Arithmetic So I've run into an issue with simplifying the statement, with ω being the first infinite ordinal:
$(ω2)^ω$
Now in the case of a finite ordinal $a$ as an exponent, it is easy to inductively prove that
$(ω2)^a = (ω^a)2$
However, this is a little unclear as to what happens with the infinite exponent.
So which of the following would be true and why:
$(ω2)^ω = (ω^ω)2$
-or-
$(ω2)^ω = ω^ω$   ?
We could consider $(ω2)^ω$ as either
$ω(2ω)(2ω)...2$
(which in some way seems like nonsense as there is no "last" term) or as
$ω(2ω)(2ω)...$
The problem is that we are oscillating back and forth between $ω$ and $2$.  Does this even have a clearly defined value?
 A: HINT: I’m assuming that by $\omega2$ you mean the ordinal product $\omega\cdot 2=\omega+\omega$. For any ordinal $\alpha$, $\alpha^\omega=\sup_{n\in\omega}\alpha^n$, so you have
$$(\omega\cdot 2)^\omega=\sup_{n\in\omega}(\omega\cdot 2)^n=\sup_{n\in\omega}\left(\omega^n\cdot 2\right)\;.$$
Clearly $\omega^n<\omega^n\cdot 2<(\omega\cdot 2)^\omega$ for each $n\in\omega$, and $\sup_{n\in\omega}\omega^n=\omega^\omega$, so $\omega^\omega\le(\omega\cdot 2)^\omega$. Try to establish the opposite inequality as well, by considering how $\omega^n\cdot 2$ compares with $\omega^\omega$.
A: In general. If $k$ is a limit ordinal then $\sup_{a<k}b^a=b^k.$
Proof: $b^a$ is defined as the order-type of the set of functions $f:a\to b$ such that $\{c\in b: f(c)\ne 0\}$ is finite. Denote this set by $(b^a)^*.$
The defined order $<^*$ on $(b^a)^*$ is reverse-lexicographic: For non-empty $a,b$ and distinct $f,g \in (b^a)^*$ let $c=\max \{d:f(d)\ne g(d)\}$. Define  $f<^*g \iff f(c)<g(c).$ 
For the non-trivial case $b\ne 0\ne k,$ let $U=\sup_{a<k}b^a=\cup_{a\in k}b^a.$
(i).  Clearly $x\in U\implies \exists a\in k\;(x\in b^a\leq b^k)\implies x\in b^k.$ So $U\leq b^k.$
(ii). To show that $U\geq b^k,$ let $\psi : [b^k,\epsilon]\to [(b^k)^*,<^*]$ be the (unique) order-isomorphism. For any $x\in b^k$ take any $a\in k$ such that $\sup \{c\in k:\psi(x)(c)\ne 0\}<a.$ Then $x$ is less than the order-type of $\{g\in (b^k)^*: \{c:g(c)\ne 0\}\subset a.\}$
But this last set is order-isomorphic to $(b^a)^*,$ which is order-isomorphic to $b^a.$  So $x<b^a$ for some $a\in k.$ That is, $x\in b^k\implies x\in U.$
In your Q, for $n<\omega$ we have $\omega^n \leq (\omega 2)^n<\omega^{\omega}.$ Therefore $\omega^{\omega}=\sup_{n<\omega}\omega ^n\leq \sup_{n<\omega}(\omega 2)^n=(\omega 2)^{\omega}\leq \omega^{\omega}.$   
