Nilpotent implies intersection of normal subgroup with upper central series is strictly ascending till the subgroup is reached

Statement
Suppose $$G$$ is a nilpotent group and $$H$$ is a normal subgroup of $$G$$. Suppose $$Z^{(i)}(G)$$ denotes the $$i^{th}$$ member of the upper central series of $$G$$, i.e., $$Z^{(0)}(G)$$ is the trivial subgroup, $$Z^{(1)}(G)$$ is the center, and $$Z^{(i+1)}(G)/Z^{(i)}(G)$$ is the center of $$G/Z^{(i)}(G)$$.

Let $$r$$ be the smallest nonnegative integer such that $$H \le Z^{(r)}(G)$$. Then $$H \cap Z^{(i)}(G)$$ is a proper subgroup of $$H \cap Z^{(i+1)}(G)$$ for all $$0 \le i < r$$.

Weaker facts

 * Nilpotent implies center is normality-large: A special case, where we only use that $$H \cap Z^{(0)}(G) \le H \cap Z^{(1)}(G)$$ for $$r \ge 1$$.

Applications

 * Normal of prime power order implies contained in upper central series member corresponding to prime-base logarithm of order in nilpotent: This says that a normal subgroup of order $$p^r$$ in a nilpotent group $$G$$ is contained in $$Z^r(G)$$.