Multiplying two ordinals where one has been raised to power of $\omega$. Term order matters?

Question

When multiplying two ordinals that are both raised to some power, my book says that one adds the exponents. But what happens if one of the exponents is $\omega$ ? Does the order of the terms matter ?

here is an example from my book:

$\omega * \omega^n = \omega^{1+n}$

Now, what if 'n' = $\omega$ ?

I am guessing that $\omega^\omega *\omega = \omega^{\omega+1}$

But is this true: $\omega^\omega *\omega = \omega * \omega^\omega$ ?

I am asking because my understanding is that adding the exponents " $\omega + 1$ " would be different from adding the exponents " $1 + \omega$ " if the rules of ordinal addition were being followed. (Since under ordinal addition rules $1 + \omega = \omega$, while
$ \omega + 1 \neq \omega$

Thanks in advance for any guidance you can provide.

Answer 1

As you mention, addition of ordinals is not commutative. And you gave the example $$\omega=1+\omega \neq \omega +1$$

Based on that you get $\omega^\omega \cdot \omega = \omega^{\omega+1}$ while $\omega \cdot \omega^\omega = \omega^{1+\omega}=\omega^\omega$. And those last two ordinals are different. More precisely $$\omega^\omega < \omega^{\omega+1}$$

Answer 2

No, we have $\omega \times \omega^\omega=\omega^\omega$.

Note that we have $\omega \times \delta < \omega^\omega$ for all $\delta <\omega^\omega$, so the limit is $\omega^\omega$ and thus gives $\omega \times \omega^\omega=\omega^\omega$.

Another way to look at it: $\omega \times \omega^\omega = \omega^{1+\omega} = \omega^\omega$ by the rule in your book.