Nilpotency of fixed class is subgroup-closed

Statement
Suppose $$G$$ is a nilpotent group and $$H$$ is a subgroup of $$G$$. Then, $$H$$ is also a nilpotent group and its nilpotency class is at most equal to the nilpotency class of $$G$$.

Note that we say that a group is nilpotent "of class $$c$$" if its nilpotency class is at most $$c$$. The statement can thus be reformulated as saying that the property of being nilpotent of class $$c$$ is closed under taking subgroups.

Proof using the upper central series definition
The idea here is to show that the upper central series ascends at least as fast as the series obtained by intersecting $$H$$ with the upper central series of $$G$$.

Proof using the lower central series definition
The idea here is to show that the lower central series descends at least as fast as the series obtained by intersecting $$H$$ with the lower central series of $$G$$.

Proof using the central series definition
The idea here is to show that intersecting with $$H$$ every member of any central series of $$G$$ gives a central series of $$H$$.

Proof in terms of iterated commutators being trivial
The proof in terms of this definition is tautological.