= Solution
For the <Abelian> case of <Cauchy theorem for groups>, use induction on $|G|$. Choose $a\ne e$. If $p$ divides $m=|a|$, then $a^{m/p}$ has order $p$. Otherwise $\langle a\rangle$ is a nontrivial <normal subgroup> of order prime to $p$, and $G/\langle a\rangle$ has smaller order still divisible by $p$. Induction gives a coset of order $p$; any lift $b$ has order divisible by $p$, because the order of its <image> divides its own order. A suitable power of $b$ then has order $p$. This uses no classification of finite <Abelian groups>.
For a finite <p-group>, the <class equation> is
$$
|G|=|Z(G)|+\sum_{[g]\ne\{z\}}[G:C_G(g)],
$$
where the sum is over noncentral <conjugacy classes>. Each index is a positive power of $p$, hence divisible by $p$. Therefore $p$ divides $|Z(G)|$, and the <center of a group> is nontrivial.
If $|G|=p^n$ with $n>1$ and $Z(G)\ne G$, this is already a proper nontrivial <normal subgroup>. If $Z(G)=G$, the <group> is <Abelian>; the established case of <Cauchy theorem for groups> gives a subgroup of order $p$, which is normal and proper. Thus
$$
\boxed{\text{No group of order }p^n\ (n>1)\text{ is simple}.}
$$
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