FSA for $(ab)^*(cb^n)^*$

Question

How can I prove that this language is regular, possibly by making a finite automata for this: $(ab)^*(cb^n)^*$, where $n\ge1$?

An automaton can easily be drawn for the part $(ab)^*$, but the part $(cb^n)^*$ doesn't seem to be regular because if the Kleene closure is taken $\ge2$ times, then it will be of the form $(cb^n)(cb^n)$, which reduces to string matching and is context-sensitive.

This was actually a question, solution to which said that the above language has the following regular expression: $(ab)^*(cbb^*)^*$.

Answer 1

For the correct/sensible interpretation, $(ab)^*(cbb^*)^*$, I will list the states and its corresponding regular expressions below; you can figure out the transitions between the states.

$q_{0}$: $(ab)^*$, start state
$q_{1}$: $(ab)^*a$
$q_{2}$: $(ab)^*(cbb^*)^*c$
$q_{3}$: $(ab)^*(cbb^*)^*$, accepting state
$q_{4}$: for all other strings, rejecting state/dead-end state

For the incorrect/unlikely interpretation, where $(cb^n)^*$ means all words like the empty word, $cb^n$, $cb^ncb^n$ and $cb^ncb^n\cdots cb^n$ for some $n\ge1$, the corresponding language is not regular since its intersection with the language with regular expression $cb^*cb^*$, $\{cb^ncb^n\mid n\ge1\}$ is not a regular language. So there is no FSA for that language.