Let \(p\) be any prime number and let \(n \ge 2\) be an integer.
Let \(G\) be a group with order \(p^n\). We show that \(G\) is not simple.
As \(G\) is a \(p\)-group, \(Z(G) \neq (1)\). (See here for proof.)
If \(Z(G) \neq G\), then \(Z(G)\) is a proper nontrivial normal subgroup of \(G\) and thus, \(G\) is not simple.
Now, assume \(Z(G) = G\). (This means that \(G\) is abelian.)
Let \(1 \neq x \in G\). Then, ord\((x) = p^m\) for some \(1 \le m \le n\).
Choose \(y = x^{p^{m-1}}\). Then ord\((y)=p\). (How?)
Let \(H = \langle y \rangle\). Then, \(H\) is a proper nontrivial subgroup of \(G\) which is normal as \(G\) is abelian.