Are Wieferich primes Wieferich numbers?

Question

In the article 'On a conjecture of Crandall concerning the $qx+1$ problem' by Franco and Pomerance, they define Wieferich primes to be prime numbers $p$ for which $p^2|2^{p-1}-1$ and Wieferich numbers to be odd positive integers $q$ for which $q^2|2^{l(q)}-1$, where $l(q)$ is the order of $2$ in the group $(\mathbb{Z}/q\mathbb{Z})^*$. They then mention that Wieferich primes are Wieferich numbers. But I don't see how $p^2|2^{p-1}-1$ implies that $p^2|2^{l(p)}-1$. It would make sense if it was $\varphi(q)$ instead of $l(q)$, which is an alternative definition, but this is not the one they gave. Is their statement true?

Answer 1

$l(p)$ is the least positive number $l$ that $p|2^l-1$.

Assume $p^2\nmid 2^l-1$. Then, $p|2^{l(\frac{p-1}{l}-1)}+...2^l+1\Rightarrow p|1+...1=\frac{p-1}{l}$ since $2^l\equiv 1$ mod $p$ and this leads to contradiction.