How to prove that concatenating a language A and A* is commutative? Suppose we have a language $A$. I want to prove that $AA^*$ is commutative. I know that this expression equals $A^+$, but I'm not sure how to go about a proof yet. This is my attempt so far.
If $A$ is a language, then
\begin{align*}
AA^* &amp; = A(A^0\cup A^1\cup\cdots) \tag{Definition of $A^*$}\\
     &amp; = A^1\cup A^2\cdots        \tag{Distributive law}   \\
     &amp; = (A^0\cup A^1\cup\cdots)A \tag{Distributive law}  \\
     &amp; = A^*A.
\end{align*}
Is this enough? I'm thinking that in order for it to be a proper proof, I have to show that $AA^*$ and $A^*A$ are subsets of each other. That is, take an element $s = xy$ where $x\in A$ and $y\in A^*$ and show that $x\in A^*$ and $y\in A$ (and vice versa). However, I get stuck very early when doing that method.
$AA^*$ is the set of all strings consisting of a string in A, followed by zero or more strings in A, which means all strings consisting of one or more strings in A. $A^*A$ is the set of all strings consisting of zero or more strings in A, followed by one string in A, which also means all strings consisting of one or more strings in A. So both are the same.