Ordinal $10^\omega$ $10^\omega$ = $10 \cdot 10 \cdot 10 \cdot ...= \lim_{\alpha \lt \omega} (10^\alpha) = \omega$. Are my thoughts correct? Is this sufficient explanation, given the ordinal arithemtic proved from ZFC?

I'd say it would be better if you referred to the appropriate definition of ordinal arithmetic. 
For an ordinal $\alpha$ and a limit ordinal $\beta$ (like e.g. $\beta = \omega$) exponentiation is defined as
$\alpha^\beta = \bigcup_{\gamma < \beta} \alpha^\gamma$
Replacing the appropriate values in the definition we therefore have:
$10^{\omega} =  \bigcup_{n < \omega} 10^n$
Since for $n \le m$ we have the set inclusion $10^n \subseteq 10^m \subseteq \omega$ it should now become apparent that $\bigcup_{n < \omega} 10^n = \bigcup_{n < \omega} n = \omega$.
