We can also define $a_0 = A$ and extend $a_{n+1} = A^{a_n}$ to all $n\geq 0$. For $A>1$ the sequence $(a_n)_{n\geq 0}$ is clearly (strictly) increasing. Therefore $2^{a_n} > 2^{a_0} = 2^A \geq 2A > A$ for all $n\geq 1$. Let us now rather prove that $b_n \geq Aa_n$ for each $n$. This is true for $n=1$, since $b_1 = A^{A+1} = A\cdot A^A = Aa_1$. Then, by the induction hypothesis $$b_{n+1} = 2^{b_n} \geq 2^{Aa_n} = (2^A)^{a_n} \geq (2A)^{a_n} = 2^{a_n}\cdot A^{a_n} > Aa_{n+1}.$$