Which ordinal is bigger, $\omega^\omega\omega$ or $\omega\omega^\omega$? I know that $\omega^\omega\omega$ is $\omega$ copies of $\omega^\omega$, and $\omega\omega^\omega$ is $\omega^\omega$ copies of $\omega$. But how can I tell which one is a member of the other?
$\omega^\omega\omega\in\omega\omega^\omega$ or $\omega\omega^\omega\in\omega^\omega\omega$? And why?
 A: The ordinal $\omega^\omega\cdot\omega$ is equal to $\omega^{\omega+1}$. 
The ordinal $\omega\cdot\omega^\omega$ is just equal to $\omega^\omega$. 
So $\omega^\omega\cdot\omega$ is the bigger ordinal while $\omega\cdot\omega^\omega$ is the smaller ordinal.
One way of understanding $\omega.\omega^\omega$ might be that: 
After $\omega$ copies of $\omega$ are placed the ordinal reached is $\omega^2$. After $\omega ^2$ copies of $\omega$ are placed the ordinal reached is $\omega^3$. After $\omega ^3$ copies of $\omega$ are placed the ordinal reached is $\omega^4$. After $\omega ^4$ copies of $\omega$ are placed the ordinal reached is $\omega^5$. And so on... 
So when we have placed $\omega^ \omega$ copies of $\omega$ we are just at $\omega^ \omega$. And not surprisingly this is supremum (least-upper-bound) of :
$$\{\,\omega ^2,\omega ^3,\omega ^4,\omega^5,...\}$$
Edit:
Note that partly why this kind of question may arise is that for any finite ordinal $n$ we actually have:
$$\omega\cdot\omega^n=\omega^n\cdot\omega=\omega^{n+1}$$
