How is $L$ = {$ { a^nsb^n : s \in L(a^*b^*) ,\ n \ge 1 } $} a regular language?
You don't need to "keep track of $n$ being the same." Another way to write the language is

$$L = \{a^na^ib^jb^n\mid n\geq 1,\ i,j\geq 0\}=\{a^{n+i}b^{j+n}\mid n\geq 1,\ i,j\geq 0\}\,.$$

So, now, given a string $a^sb^t$, you just need to decide if you can choose $n\geq1$ and $i,j\geq 0$ such that $s=n+i$ and $t=n+j$.  If so, then $a^sb^t\in L$; otherwise, $a^sb^t\notin L$. If $s=0$ or $t=0$, we can't choose $n$, $i$ and&nbsp;$j$ as required, since $n\geq 1$. However, as long as $s&gt;0$ and $t&gt;0$, we can take $n=1$, $i=s-1$ and $j=t-1$. Therefore,

$$L = \{a^sb^t\mid s,t\geq 1\}\,,$$

which is the language matched by the regular expression $aa^*b^*b$.