Can I choose a positive real number $a\in\Bbb R^+$ so that $1,a,a^a,a^{a^a},...$ are independent in the sense that no combination of integer coefficients will add up these numbers to zero?
More formally: Given $A_0=1$ and $A_{i+1}=a^{A_i}$. Is there an $a\in\Bbb R^+$ for which I can be sure that there is no $k\in\Bbb N^+$ and no coefficients $n_i\in\Bbb Z,i=0,1,...,k\;$ with
$$n_0A_0+n_1A_1+\cdots +n_kA_k=0\quad?$$
I believe such numbers exists, and are actually very common. I just have no clue how to prove it.
For each particular linear combination, $$n_0A_0+n_1A_1+\cdots +n_kA_k$$ is an analytic function of $a$ in $(0,\infty)$. Since it is not identically zero, the set of its zeroes is discrete. In particular this set (which is the $a$s that are excluded by that particular linear combination) has Lebesgue measure zero.
Since there are only countably many integer combinations, this means that the union of the sets they exclude also has measure zero.
This means that there are plenty of $a$s that are not excluded -- in fact almost every $a$ has an independent sequence of tetrations.