Proving that a (tree) language is not Buchi recognizable
The idea is to play with the $m_i$'s, as follows.

Consider a tree $t_1$, which has $a$ in $\epsilon$, and nowhere else. Look at the path $1^\omega$ &mdash; an accepting run on it has infinitely many accepting states. Wait until such an accepting state was reached, and only then put $a$ at $1^{m_1}0$, where $m_1$ is large enough.

Now look at the path $1^{m_1}01^\omega$, again, it has an accepting state eventually, denote the length up to it by $m_2$, put an $a$ at $1^{m_1}01^{m_2}0$ and look at $1^{m_1}01^{m_2}01^\omega$. By now you have almost what you wanted - two accepting states with an $a$ between them (informally speaking). But you are not yet guaranteed that both accepting states are equal. To enforce the latter, we repeat the process for $n$ times, thus exhausting the number of accepting states, and by the pigeonhole principle, two accepting states must be equal.