Finite upper central series members are purely definable

This article gives the statement, and possibly proof, of the fact that for any group, the subgroup obtained by applying a given subgroup-defining function always satisfies a particular subgroup property (i.e., purely definable subgroup)}
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Statement

Suppose $G$ is a group. Let $Z^{k}(G)$ be the (finite) Upper central series (?) of $G$ . Explicitly, $Z^{k}(G)$ for all nonnegative integers $k$ is defined as follows:

$Z^{0}(G)$ is the trivial subgroup
$Z^{k}(G)$ is the subgroup containing $Z^{k-1}(G)$ such that $Z^{k}(G)/Z^{k-1}(G)$ is the center of $G/Z^{k-1}(G)$ .

Then, $Z^{k}(G)$ is a purely definable subgroup of $G$ for each $k$ .

Facts used

Proof

Direct proof

We have $g\in Z^{k}(G)$ if it satisfies the following sentence $\varphi (g)$ :

$\varphi (g)=\ \forall \ x_{1},x_{2},\dots ,x_{k}\in G,\ :\ [[\dots [g,x_{1}],x_{2}],\dots ,x_{k}]=e$

where $e$ is the identity element.

Proof using given facts

The proof follows by induction using facts (1) and (2). PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]