I am trying to prove that every $p$-group is solvable, i.e., $G$ is solvable if $|G|=p^k$ for a prime $p$ and integer $k$.
I want to show that if the commutator subgroup $G^{(1)}$ is strictly smaller than $G$, then we can do an induction since $|G^{(1)}|=p^l$ for some $l<k$ is another $p$-group, and we keep reducing the exponent as we take further commutator subgroups until it becomes 1, i.e., $|G^{(n)}|=p^0=1$, which tells $G$ is solvable.
I know the commutator group tells how "non-abelian" a group is, so I think if there is at least some abelian property in $G$, the commutator subgroup would be strictly smaller than itself. Then I recall a result that every $p$-group has a nontrivial center $Z(G)$. However, I get stuck here in proving that if $Z(G)$ is nontrivial, then the commutator subgroup is strictly smaller, as in the title.
Is the statement true? How to prove it if so? If it's false, then how to prove the original proposition that every $p$-group is solvable?