we can write $A$ as $A=\bigcup_{i=1}^m A_i$ with $m{\!\leq\!}A_1{\!<\!}A_2{\!<\!}\ldots{\!<\!}A_n$ and $A_i{\!\in\!}{\mathcal S}_\beta[{\mathcal S}_{\gamma_1}]$, for $i{\!=\!}1,\ldots, n$ which in turn means that $A_i{\!=\!}\bigcup_{j=1}^ {m_i} A_{(i,j)}$, where $A_{(i,1)}{\!<\!}A_{(i,2)}{\!<\!}\ldots {\!<\!}A_{(i,l_i)}$, $A_{(i,j)}{\!\in\!}{\mathcal S}_{\gamma_1}$, for $j{\!=\!}1,2,\ldots, l_i$, and $\{\min(A_{(i,j)}:j=1,2,\ldots,l_i\}\in {\mathcal S}_\beta$, for $i=1,2,\ldots , m$. This means that $\{ \min A_{(i,j)}:j=1,2,\ldots,l_i, i=1,2,\ldots, m\}$ is in ${\mathcal S}_{\beta+1}$ and thus we conclude that $A\in {\mathcal S}_{\beta+1}[{\mathcal S}_{\gamma_1}]$. Conversely, we can show in a similar way that $ {\mathcal S}_{\beta+1}[{\mathcal S}_{\gamma_1}]\subset  {\mathcal S}_1\big[{\mathcal S}_\beta[{\mathcal S}_{\gamma_1}]\big] $.

If $\gamma_2$ is a limit ordinal we first observe that $$\lambda(\gamma, n)= \lambda({\gamma_1+\gamma_2},n )=\gamma_1+ \lambda(\gamma_2,n ).$$ If $A\in {\mathcal S}_{\gamma_1+\gamma_2}$ it follows that there is an $n\le \min A$ so that, using the induction hypothesis, we have $$A\in  {\mathcal S}_{\gamma_1+\lambda(\gamma_2,n)}={\mathcal S}_{\lambda(\gamma_2,n)}[{\mathcal S}_{\gamma_1}].$$ This means that $A=\bigcup_{j=1}^m A_j$ with, $A_1<A_2<\ldots< A_m$, $\{\min(A_j):j=1,2,\ldots, m\}\in {\mathcal S}_{\lambda(\gamma_2,n)}$ and $A_j\in{\mathcal S}_{\gamma_1}$, for $j=1,2,\ldots, m$. Since $n\le\min(A)=\min(A_1)$, it follows that $\{\min(A_j):j=1,2,\ldots,m\}\in {\mathcal S}_{\gamma_2} $, and, thus, that $A\in {\mathcal S}_{\gamma_2}[{\mathcal S}_{\gamma_1}]$. Conversely, we can similarly show that if $A\in   {\mathcal S}_{\gamma_2}[{\mathcal S}_{\gamma_1}]$, then it follows that $A\in{\mathcal S}_{\gamma_1+\gamma_2}$.

If Case 1 does not hold $\alpha$ must be of the form $\alpha=\omega^\gamma$.

Case 2. $\alpha=\omega^{\omega^\kappa}$, for some $\kappa<\omega_1$. In that cases we make the following requirement on 