If $\alpha$ is any algebraic number, there is an integer $d > 0$ such that $d\alpha$ is an algebraic integer, and the minimum such $d$ is called the denominator of $\alpha$, written $\text{den}(\alpha)$.
Now consider a sequence of algebraic integers $(a_i)$ such that $$ \alpha = \prod_{i = 1}^{\infty} a_i $$ is algebraic. I'm wondering if there are some nice sufficient conditions for the existence of an index $n$ such that $\alpha = a_n \alpha'$ with $\text{den}(\alpha) = \text{den}(\alpha')$. It would also be nice to know if we can choose $n$ such that $a_n \mathcal{O}_K$ and $\text{den}(\alpha) \mathcal{O}_K$ are coprime, where $K = \mathbb{Q}(a_n)$.
I tried to find some examples of $\alpha$ and play with those, but so far I came up short. I can see that the answer to both questions is positive for $$ n = n^{\sum_{i = 1}^{\infty} 1/2^i} = \prod_{i = 1}^{\infty} n^{1/2^i} $$ with $n$ any fixed positive integer, but I couldn't even extend this to $$ \alpha = \prod_{i = 1}^{\infty} n^{1/m_i} $$ for any given sequence of integers $(m_i)$ such that the product converges. In general I only know that if $\alpha$ is irrational and algebraic, then $\sum_{i = 1}^{\infty} 1/m_i$ cannot be in $\overline{\Bbb{Q}} \setminus \Bbb{Q}$ (as a consequence of Baker's theorem).
Motivation: Such a factorisation would imply a neat proof that some numbers, like $\prod_{k=2}^{\infty} k^{1/2^{k-1}}$, are transcendental.
