Solution
= Solution
{c}
The <multiplicative group of a finite field> $p$ has <group order> $p-1$: its elements are exactly the nonzero residue classes. For any <prime factor> $p$ of $2^q-1$, <6/c/i> and <Lagrange's theorem> therefore imply
$$
q=\operatorname{ord}_p(2)\mid p-1.
$$
Since $q$ and $p-1$ are positive <integers>, $q\leq p-1$, and hence
$$
\boxed{p>q.}
$$
This applies to every <prime factor>, including repeated <prime factors> viewed as the same <prime number>.