Fibonacci numbers and prime numbers I'm sorry but I couldn't find anything similar and I don't know how to search my problem so I ask it. I hope that is not duplicate.
If we call $n$th element of Fibonacci sequence with $F_n$ and if $p$ is a prime number (except 5). we want to prove that:
$GCD(p, F_p) = 1$
$GCD(p, F_{F_p}) = 1$
$GCD(p, F_{F_{\cdots_{F_{p}}}}) = 1$
Is it possible to help me? Thanks.
I'm sorry for bad English too.  
 A: For the first of your conjectures, which can be reformulated as '$p$ does not divide $F_p$' (see the comments by TheSimpliFire), the proof can be found on Wikipedia  (in somewhat cryptic form). The wiki-page on Fibonacci primes states:

A prime $p$ divides $F_{p-1}$ if and only if $p$ is congruent to ±1 modulo 5, and $p$ divides $F_{p+1}$ if and only if is congruent to ±2 modulo 5. (For $p = 5$, $F_5 = 5$ so $5$ divides $F_5$)

Now suppose that $p \neq 5$ and $p | F_p$. We will derive a contradiction. We know that either $p = \pm 1 \mod 5$ or $p = \pm 2 \mod 5$. We look at both cases separately. If $p = \pm 1 \mod 5$ then by Wikipedia $p|F_{p-1}$. Also, by assumption, $p|F_p$. It follows that $p|(F_{p-1} + F_p)$. In other words: $p|F_{p+1}$. But this means by Wikipedia that $p = \pm 2 \mod 5$, a contradiction. Similar, if $p = \pm 2 \mod 5$ we have that $p|F_{p+1}$ and $p|F_p$ so that $p|F_{p-1}$ which gives $p = \pm 1 \mod 5$, contradiction.
I don't know how to do the other conjectures, though. $GCD(p, F_p) = 1$ gives $GCD(F_p, F_{F_p}) = 1$ but that gives no or very little information on $GCD(p, F_{F_p})$
