Yuval showed a simple way to prove this. Here's an (arguably more complex) alternative based on least fixed points.
The inclusion $L \subseteq L^*$ always holds, so $\mathcal A^* \subseteq \mathcal A^{**}$.
We are left with proving $\mathcal A^{**} \subseteq \mathcal A^{*}$. For this, recall that $L^*$ can be defined as the least language such that
$$
\{\epsilon\} \cup LL^* = L^*
$$
Hence, $\mathcal A^{**}$ is the least language such that
$$
\{\epsilon\} \cup \mathcal A^* \mathcal A^{**} = \mathcal A^{**}
$$
so, if we prove that $\mathcal A^*$ also satisfies the same property, i.e. if we prove
$$
\{\epsilon\} \cup \mathcal A^* \mathcal A^{*} = \mathcal A^{*}
\qquad (*)
$$
by the minimality of $\mathcal A^{**}$, we will obtain the wanted
$\mathcal A^{**} \subseteq \mathcal A^{*}$.
Proving $(*)$ is then trivial.