Solution: Since the group has at least two elements, take $g \in G$ which is not identity. Since $G$ is a finite group, $g^n = e$ for some $n \in \mathbb{N} $. Further assume that $n$ is the order of $g$. If $n$ is a prime, we are done. If not, let $p$ be a prime factor of $n$, so $ n = pk$ for $k \in \mathbb{N} $. Thus, \[ e = g^n = g^{pk} = \left( g^k \right) ^p. \] Let $h = g^k$, so we have $h^p = e$. Now we claim that the order of $h$ is $p$. It is clear that $\mathrm{ord}(h) \leq p $. If $\mathrm{ord} (b) = m < p$, then \[ e = b^m = \left( a^k \right) ^m = a^{km}, \] which is a contradiction to $ord(a) = n > km$. Hence, $\mathrm{ord} (b) = p$.