If a nilpotent group has an element of prime order $p$, so does its centre.

It is also not too hard to prove this statement directly by induction on the nilpotency class of $G$:

If the nilpotency class is 0 or 1, there is nothing to show, as $G$ is abelian in this case. So let the class of $G$ be greater than 1, and let $a \in G$ be any element of order $p$. If $a$ is in $Z(G)$, we are done. Otherwise, $a$ leads to an element of $G/Z(G)$ of order $p$, and by induction hypothesis we know that $G/Z(G)$ has an element of order $p$ in its centre. In other words, there is an element $b \in Z_2(G)$ such that $b \notin Z(G)$ and $b^p \in Z(G)$. Since $b$ is not in the centre of $G$, there is some $c \in G$ such that the commutator $[b,c] \in Z(G)$ is nontrivial. Now we can complete the proof by showing $[b,c]^p = [b^p,c] = 1$.

In the proof of 5.2.7 is proven that, for all primes $p$, $G$ has a $p$-subgroup containing all $p$-subgroups of $G$, say $K(p)$. Then $K(p)$ is clearly a characteristic subgroup of $G$, in particular, $K(p)$ is a normal subgroup of $G$. If $G$ has an element of order $p$, $p$ a prime, then $K(p)\neq1$. But $G$ is nilpotent and so there exists $i\geqslant1$ such that $$[K(p),\underbrace{G,\ldots,G}_{i}]=1.\tag{1}$$ If $K(p)\leq Z(G)$ we get the result. On the other hand, if $K(p)\nleq Z(G)$, then let $i_0$ be the least positive integer such that $(1)$ holds. Thus $i_0>1$ and $$H=[K(p),\underbrace{G,\ldots,G}_{i_0-1}]\neq1.$$ Finally, $H\leq Z(G)$ since $[H,G]=1$.