Closure properties of Alternating Circuit 1 level
Denote your language by $L$. Let the input be $x_1,\ldots,x_n$. Let $f(i,j,\ell)$ be true if $x_1,\ldots,x_j \in L^\ell$. The output will be
$$
\bigvee_{\ell=1}^n f(1,n,\ell).
$$
When $\ell = 1$, there is an $\mathsf{AC^1}$ circuits for $f(i,j,\ell)$. When $\ell &gt; 1$, we use the formula
$$
f(i,j,\ell) = \bigvee_{k=i}^{j-1} (f(i,k,\lfloor \ell/2 \rfloor) \land f(k+1,j,\lceil\ell/2\rceil)).
$$
The resulting circuit is still $\mathsf{AC^1}$, since there are only $O(\log n)$ levels of recursion.
A different way of looking at this construction is to notice that for inputs of length $L$, we can replace $L^*$ with $L \cup \cdots \cup L^n$; and that $L^m$ can be computed in $O(\log m)$ steps using repeated squaring.