 Solution

Suppose there were only finitely many prime numbers. There is at least one, namely $2$, so this finite set would have a largest element $q$. The integer $2^q-1$ is greater than $1$ and therefore has a prime factor $p$: for example, its least divisor greater than $1$ must be a prime number, since any proper factor of that divisor would contradict minimality.

By ii, $p>q$, contradicting the choice of $q$. Therefore there are infinitely many prime numbers, proving Euclid's theorem by orders of group elements in finite multiplicative groups.