Proper-power language is not regular
= Proper-power language is not regular
Over an alphabet containing $a,b$, the language $\{w^n:n\geq2\}$ is not <regular language>. Intersecting it with the regular language $a^+ba^+b$ gives exactly
$$
\{a^mba^mb:m\geq1\},
$$
which fails the <pumping lemma for regular languages>.