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The answer seems to be affirmative. We use the idea of Henry Wilton that the image might be taken as an alternating group $A_q$, a simple one. Let $K=\mathcal A(1).$ Then

$\mathcal A(m) \ge [K,K,\ldots,K]=[..[K,K],..,K]\qquad (m\quad times) \qquad (*)$

Take a nontrivial $\alpha \in [K,K]$ and a surjective homomorphism $\Delta: \mathrm{Aut}(F_n) \to A_q$ which doesn't vanish at $\alpha$.

Then $ A_q =\mathrm{NormalClosure}(\Delta(\alpha))=\Delta([K,K])=\Delta(K). $

The answer seems to be affirmative. We use the idea of Henry Wilton that the image might be taken as an alternating group $A_q$, a simple one (see his comment above). Let $K=\mathcal A(1).$ Then

$\mathcal A(m) \ge [K,K,\ldots,K]=[..[K,K],..,K]\qquad (m\quad times) \qquad (*)$

Take a nontrivial $\alpha \in [K,K]$ and a surjective homomorphism $\Delta: \mathrm{Aut}(F_n) \to A_q$ which doesn't vanish at $\alpha$.

Then $$ A_q =\mathrm{NormalClosure}(\Delta(\alpha))=\Delta([K,K])=\Delta(K). $$
It follows that $$ \Delta( [K,K,\ldots,K])=A_q $$ and by $(*)$ $\Delta( \mathcal A(m))=A_q $ for every $m \ge 1.$

The answer seems to be affirmative. We use the idea of Henry Wilton that the image might be taken as an alternating group $A_q$, a simple one. Let $K=\mathcal A(1).$ Then

$\mathcal A(m) \ge [K,K,\ldots,K]=[..[K,K],..,K]\qquad \text{($m$ times)} \qquad (*)$

Take a nontrivial $\alpha \in [K,K]$ and a surjective homomorphism $\Delta: \mathrm{Aut}(F_n) \to A_q$ which doesn't vanish at $\alpha$.

Then $ A_q =\mathrm{NormalClosure}(\Delta(\alpha))=\Delta([K,K])=\Delta(K). $

The answer seems to be affirmative. We use the idea of Henry Wilton that the image might be taken as an alternating group $A_q$, a simple one. Let $K=\mathcal A(1).$ Then

$\mathcal A(m) \ge [K,K,\ldots,K]=[..[K,K],..,K]\qquad (m\quad times) \qquad (*)$

Take a nontrivial $\alpha \in [K,K]$ and a surjective homomorphism $\Delta: \mathrm{Aut}(F_n) \to A_q$ which doesn't vanish at $\alpha$.

Then $ A_q =\mathrm{NormalClosure}(\Delta(\alpha))=\Delta([K,K])=\Delta(K). $

The answer seems to be affirmative. We use the idea of Henry Wilton that the image might be taken as an alternating group $A_q,$ a simple one. Let $K=\mathcal A(1).$ Then $$ \mathcal A(m) \ge [K,K,\ldots,K]=[..[K,K],..,K]\qquad \text{($m$ times)} \qquad (*) $$ Take a nontrivial $\alpha \in [K,K]$ and a surjective homomorphism $\Delta: \mathrm{Aut}(F_n) \to A_q$ which doesn't vanish at $\alpha.$ Then $$ A_q =\mathrm{NormalClosure}(\Delta(\alpha))=\Delta([K,K])=\Delta(K). $$ It follows that $$ \Delta( [K,K,\ldots,K])=A_q $$ and by $(*)$ $$ \Delta( \mathcal A(m))=A_q. $$ for every $m \ge 1.$

The answer seems to be affirmative. We use the idea of Henry Wilton that the image might be taken as an alternating group $A_q$, a simple one. Let $K=\mathcal A(1).$ Then

$\mathcal A(m) \ge [K,K,\ldots,K]=[..[K,K],..,K]\qquad \text{($m$ times)} \qquad (*)$

Take a nontrivial $\alpha \in [K,K]$ and a surjective homomorphism $\Delta: \mathrm{Aut}(F_n) \to A_q$ which doesn't vanish at $\alpha$.

Then $ A_q =\mathrm{NormalClosure}(\Delta(\alpha))=\Delta([K,K])=\Delta(K). $