**Step 1: Choose a non-identity element.** Since $|G| = p \geq 2$, there exists $g \in G$ with $g \neq e$. **Step 2: Determine the order of $g$.** By [Element Order Divides Group Order](/theorems/783), $o(g) \mid |G| = p$. Since $p$ is prime, the only positive divisors of $p$ are $1$ and $p$. Since $g \neq e$, $o(g) \neq 1$. Therefore $o(g) = p$. **Step 3: Conclude.** Since $o(g) = p = |G|$, the cyclic subgroup $\langle g \rangle$ has order $p$ and is contained in $G$, so $\langle g \rangle = G$. Therefore $G$ is cyclic.