$\log \log (\alpha)$ transcendental?? $\log \log (\alpha)$ transcendental?? ($\alpha$ algebraic $\neq 0$ and $1$)
I supposed $\log \log (\alpha)=\beta$ , $\beta$ transcendental. Then $\log(\alpha)=e^{\beta}$ and it is know $e^{\beta}$ is transcendental.
 A: I don't know if we can prove it right now, but it is likely.
An interesting conjecture by Schanuel states that

If $x_1,\dotsc,x_n$ are complex numbers linearly independent over $\Bbb{Q}$, then
  $$
\text{trdeg}_{\Bbb{Q}}(x_1,\dotsc,x_n,e^{x_1},\dotsc,e^{x_n}) \geq n
$$

If this conjecture holds then
$$
2 \leq \text{trdeg}_{\Bbb{Q}}(\alpha, \log(\alpha), \log(\alpha), \log\log(\alpha)) = \text{trdeg}_{\Bbb{Q}}(\log(\alpha), \log\log(\alpha)) \leq 2
$$
so $\log(\alpha)$ and $\log\log(\alpha)$ would be algebraically independent for every $\alpha \in \bar{\Bbb{Q}} \setminus \{0,1\}$.
A: The answer is NOT. We are sure $\log\alpha$ is trascendental because $\alpha$ is algebraic neither $0$ nor $1$ but the log of a trascendental may be either algebraic or trascendental ($\log e=1$ and $e$ is trascendental) 
A: lindemann weirstrauss theorem states that $$ e^x$$ is transcendental For x is algebraic $$(x)$$ not equal to 0 1.
If x is transcendental then $$e^x$$ is transcendental or Algebraic. x=e^log_e(x) here x is algebraic so $$log_e(x) $$ is transcendental . Now $$log_e(x)$$=e^ln(ln(x)) here ln(x) is transcendental so ln(ln(x)) is algebraic or transcendental. Read possibly transcendental numbers on wikipeidia.
