Product of Commuting Elements in Monoid is Unit iff Each Element is Unit/Proof 2

Proof
If $x_1, \ldots, x_n \in \map G A$, then:
 * $\ds \prod_{i \mathop = 1}^k x_i \in \map G A$

by Inverse of Product: Monoid: General Result.

Conversely, suppose:
 * $\ds \prod_{i \mathop = 1}^k x_i \in \map G A$

That is, there is a $z \in A$ such that:
 * $(1):\quad \ds z \paren {\prod_{i \mathop = 1}^k x_i} = \paren {\prod_{i \mathop = 1}^k x_i} z = e$

We shall show:
 * $\forall i \in \set {1, \ldots, n} : x_i \in \map G A$

It suffices to show this for $i=1$, since $x_1, \ldots, x_n$ are commuting.

Define:
 * $\ds z_1 := \paren {\prod_{i \mathop = 2}^k x_i} z$

Then:
 * $x_1 z_1 = z_1 x_1 = e$

so that $x_1 \in \map G A$.

Indeed:

Moreover: