Proving that the language $\{ w^n\mid w \in \{0,1\}^∗, \, n \ge 2 \}$ is not regular
Suppose towards a contradiction that your language $L$ was regular and let $p$ be its pumping length.
Consider the word $w = 0^p 10^p1 \in L$. By the pumping lemma, $w$ can be written as $w=xyz$ with $x=0^{h}$, $y=0^k$, and $z=0^{p-h-k}10^p 1$ (here $h+k \le p$ and $k \ge 1$) such that, for every non-negative integer $i$, $xy^iz \in L$.
Substituting and simplifying, the above means that, for every $i \ge 0$,  we have $0^{p-k} 0^{ki} 1 0^p 1 \in L$. In particular, the above must hold for $i=0$, implying that $0^{p-k}1 0^p 1 \in L$.
Since $0^{p-k}1 0^p 1 \not\in L$ (recall that $k \ge 1$), this is provides the sought contradiction and concludes the proof.