When is the concatenation of a language $L$ with $\Sigma^*$ regular? I've been looking at questions about the regular concatenation of two languages; one question said that the concatenation of $\{0^n1^n|n\geq 0\}$ and $\Sigma^*$ was regular (over the alphabet $\Sigma = \{0,1\}$). Another provided a counterexample by the concatenation of $\{a^{n^2}b|n\geq 0\}$ with $\Sigma^*$, which is irregular (over the alphabet $\Sigma = \{a,b\}$). Could someone provide clarification as to why some concatenations with $\Sigma^*$ are regular while others aren't? Both cases concatenate an irregular language unrecognisable by a DFA to one that is, and I don't understand how this produces a language that is recognisable. 

Links to source questions: 
https://math.stackexchange.com/questions/649335/are-languages-regular-if-their-concatenation-is-regular
Why is $\Sigma^*$ concatenated with some language regular?
In the first case, $n$ can be 0, and the concatenation will produce $\Sigma^*$. So any other value of $n$ is irrelevant; no other strings could be added. If the restriction were $n&gt;1$, then the concatenation would not be regular.