= Solution
{c}
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 <6/c/ii>, $p>q$, contradicting the choice of $q$. Therefore \b[there are infinitely many <prime numbers>], proving <Euclid's theorem> by <orders of group elements> in finite multiplicative <groups>.
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