Nilpotent of cube-free order implies abelian

Statement

Suppose $G$ is a 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 Finite abelian group (?).

Related facts

• Square-free implies solvability-forcing

Facts used

1. Classification of groups of prime-square order, showing that any group of prime-square order is abelian.
2. 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.