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If $n$ is fixed then $a^n$ is just a single word and so is $a^nb^n$. If by $a^n$ you mean the language $\{a^n \mid n \ge 0\}$ (whose corresponding regular expression is $a^*$) then the problem is that the variable $n$ in $a^n$ is independent of the variable with the same name in $b^n$. The language obtained by concatenating $A=\{a^n \mid n \ge 0\}$ with $B=\... 1 The automaton is deterministic, so any string over$\{0,1\}$has a unique path. We need at least one$1$to move from$q_1$to the component where is the accepting state. Immediately after reading$1$we are always in accepting state$q_2$. Look at the last$1$in the input. At that moment we accept in$q_2$. To return to the accepting state, only reading$0$... 2 Yes, since we can let$i$be 0. Every non-empty word$x\in L$can be expressed as "$xx^0\epsilon$". We can also let$i$be 1. Every non-empty word$x\in L$can be expressed as "$\epsilon x^1\epsilon$". The statements above are rather trivial and banal. So, the real question is, given a regular language$L$, is there some$i\geq 2\$ such ...