Nilpotent of cube-free order implies abelian

Statement
Suppose $$G$$ is a fact about::finite nilpotent group and there is no prime number $$p$$ such that $$p^3$$ divides the order of $$G$$. In other words, the order of $$G$$ is a cube-free number.

Then, $$G$$ is a fact about::finite abelian group.

Related facts

 * Square-free implies solvability-forcing

Facts used

 * 1) uses::Classification of groups of prime-square order, showing that any group of prime-square order is abelian.
 * 2) uses::Equivalence of definitions of finite nilpotent group, in particular the part that states that any nilpotent group is the internal direct product of its Sylow subgroups.