= Solution
Let
$$
L_1=\{w^n:w\in\{a,b\}^*,\ n\geq2\}.
$$
Intersect it with the <regular language>
$$
R=a^+ba^+b.
$$
Every word of $R$ contains exactly two copies of $b$. If it is a proper power, its exponent must therefore be two and its root must contain one $b$. Hence
$$
L_1\cap R=\{a^mba^mb:m\geq1\}.
$$
This language is not regular: if $p$ were a <pumping length>, pumping any nonempty substring among the first $p$ copies of $a$ in $a^pba^pb$ would change only its first $a$-block. Since regular languages are closed under intersection,
$$
\boxed{L_1\text{ is not regular}.}
$$
This is the <proper-power language is not regular>.
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