Prove that the set is infinite Let $ a_0 = a> 1$  be an integer, and for $ n \ge 0$ , define $ a_{n + 1} = 2 ^ {a_n}-1$ . Show that the set of prime divisors of the terms of the sequence $ a_n$  is infinite.
This is a problem from 38th Brazilian Mathematics Olympiad.
 A: In base $10$, a repeated decimal of any form can be represented as $\frac{x}{10^n-1}$ where $x$ is the repeated digits and $n$ is how many digits $x$ has. Since any repeated fraction can be represented this way, a repeated sequence of $9$s implied by $10^n-1$ should be divisible by any number which has divisors other than $2$ and $5$ if you put in enough $9$s.
You can apply this same logic to base $2$, see if you get the answer...
A: Assume to the contrary that the set of primes is finite. Then there must exist a prime $p$ such that $p$ divides infinitely many elements of the sequence. Pick such a prime.
Suppose that $p\mid a_n=2^{a_{n-1}}-1$ with $n>>p$. Let $\ell_1$ be order of $2$ modulo $p$. Thus we must have $\ell_1\mid a_{n-1}=2^{a_{n-2}}-1$. Let $\ell_2$ be the order of $2$ modulo $\ell_1$. Then we also must have $\ell_2\mid a_{n-2}$. Continuing this we must have, $\ell_k\mid a_{n-k}$. Observe that $\ell_i\ne 1$, otherwise we will have $\ell_{i-1}\mid 1$, a contradiction. Thus this process will continue until we have $\ell_i=2$. But we also should have $2=\ell_i\mid a_{n-i}$. But $a_{n-i}$ is odd, contradiction! $\square$
