Proof of The Iterated Operation over a Finite Set Does Not Depend on the Enumeration

Nonempty. By The Number of Elements of a Finite Set §cardinality, $n\in\mathbb{N}_{0}$ and there is a bijection $\varphi$ from $[n]$ onto $A$. If $n=0$, then $[n]=\emptyset$ by Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting §segment, so $A=\varphi([n])=\emptyset$, contrary to the assumption. Hence $n\neq0$, and $n\in\mathbb{N}$ by The Natural Numbers with Zero and Their Embedding into the Integers §naturals.

Independent. Let $\varphi$ and $\psi$ be bijections from $[n]$ onto $A$; here $n\in\mathbb{N}$ by the first part. By Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §inverse, Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §composition and Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §preservation, $\sigma=\varphi^{-1}\circ\psi$ is a bijection from $[n]$ onto $[n]$ with $\varphi(\sigma(k))=\psi(k)$ for all $k\in[n]$. Put $a_{k}=f(\varphi(k))$; then $a_{\sigma(k)}=f(\psi(k))$, and since $\ast$ is associative and commutative, Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §reordering gives

$$\mathop{\ast}\limits_{k=1}^{n}f(\psi(k))=\mathop{\ast}\limits_{k=1}^{n}a_{\sigma(k)}=\mathop{\ast}\limits_{k=1}^{n}a_{k}=\mathop{\ast}\limits_{k=1}^{n}f(\varphi(k)).$$