Infinity of primes

Suppose that $p$ is the largest prime. Then $2^p-1$ is composite and let $q$ be some prime dividing $2^p-1$. Since $p$ is a prime, the order of $2$ in the multiplicative group $\mathbb{Z}_q^{\times}$ is $p$. Since $q$ is a prime, the order of $\mathbb{Z}_q^\times$ is exactly $q-1$. By Lagrange’s theorem, $p \mid (q-1)$, which implies that $p < q$.