Find ordinal numbers $a$, $b$ and $c$ so that $$(a\cdot b)^c \ne a^c \cdot b^c$$ My example would be $a=2$, $b=\omega$ and $c=2$, where $\omega$ is the ordinal number of the set of natural numbers. I know that $2\cdot\omega=\omega$ because there exists an increasing bijection between $\mathbb N$ and $\{a,b\}\times\mathbb N$ with lexicographic order. So I would have to prove that $\omega^2$ is not equal to $(2^2)\cdot(\omega^2)$, but it kind of seems to me like they are similar sets, the set on the left seems just a bit prolonged.
( two sets are similar if there exists an increasing bijection between them , and two well ordered sets have the same ordinal numbers if they are similar)
It's correct that $\omega^2 = 4\cdot\omega^2$, so your example doesn't work.
How about the other way around, though? Ordinal multiplication is associative, so
$$ (\omega\cdot 2)^2 = \omega\cdot 2\cdot \omega\cdot 2 = \omega^2\cdot 2 $$ whereas $$ \omega^2\cdot 2^2 = \omega^2\cdot 4 $$
(You then need to argue that $\omega^2\cdot 2\ne \omega^2 \cdot 4$, of course, but that should be easier because it's actually true).