A Product Recursion Meets an AP

Let $\{a_n\}$ be an arithmetic sequence with nonzero common difference and $a_5 = 6$, and let $\{b_n\}$ satisfy $b_1 = 3$ and

1. For $n \geqslant 2$, prove that $\dfrac{b_{n+1} - 1}{b_n - 1} = b_n$. 2. Suppose $a_3 > 1$ and $a_3 \in \mathbb{N}^*$, and there exist geometric sequences $a_3, a_5, a_{k_1}, a_{k_2}, \ldots, a_{k_n}$ of arbitrarily many terms drawn from $\{a_n\}$. Find $a_3$. 3. Under the conditions of part 2, with $a_3$ at its minimum value, prove that