Use this sequence to prove that there are infinitely many prime numbers. 
Problem:
By considering this sequence of numbers
$$2^1 + 1,\:\: 2^2 + 1,\:\: 2^4 + 1,\:\: 2^8 +1,\:\: 2^{16} +1,\:\: 2^{32}+1,\ldots$$
prove that there are infinitely many prime numbers.


I am thinking that if I can show that every pair of numbers in the sequence are relatively prime then since each has at least one prime factor this would prove the existence of infinitely many primes.
But I am new to discrete mathematics and number theory so I am not sure on how to proceed.
 A: If $2^{2^n}\equiv -1\pmod p$, then show that $2^{2^{m}}\not\equiv-1\pmod p$ for any $m<n$. 
This is because if $2^{2^m}\equiv -1\pmod p$ then: $$-1\equiv 2^{2^n}=\left(2^{2^{m+1}}\right)^{2^{n-m-1}}\equiv 1\pmod p$$
So $p=2$. But $p$ can't be $2$.
So $2^{2^n}+1$ and $2^{2^m}+1$ can't have any prime factors in common if $m\neq n$.
A: To continue the approach by @JérémyBlanc: Note that $a_n=2^{2^n}+1$ in general. Now, note that $$a_{n}-2=2^{2^{n}}-1=(2^{2^{n-1}}-1)(2^{2^{n-1}}+1)=a_{n-1}(2^{2^{n-1}}-1)=a_{n-1}(a_{n-1}-2)=a_{n-1}a_{n-2}(a_{n-2}-2)=\dots=a_{0}a_{1}a_{2}\dots a_{n}(a_{0}-2)=a_{0}a_{1}a_{2}\dots a_{n-1}$$
$$\therefore a_n-2=a_0a_1a_2\dots a_{n-1}$$
Equivalently, $$
2^{2^n}-1 = (2^{2^1} - 1) \prod_{m=1}^{n-1} (2^{2^m}+1),
$$ 
Since $a_{n}$ is odd, note that $\gcd(a_n,a_{n}-2)=1$. This implies that any divisor of $a_{n}-2$ is different from any divisor of $a_n$, thus implying for $t<n$, $\gcd(a_n, a_t)=1$.
This implies that all $a_n$ are coprime. As you have noticed, this completes our proof. 
A: Your sequence is given by $a_n=2^{2^n}+1$. You can observe that 
$(a_n-2)\cdot a_n=(2^{2^n}-1)(2^{2^n}+1)=(2^{2^n})^2-1=(2^{2\cdot 2^n})-1=(2^{2^{n+1}})-1=a_{n+1}-2$.
Hence if a prime number divides $a_n$ and $a_{n+1}$, then it divides $2$. But $a_n$ and $a_{n+1}$ are both odd, so they are coprime.
You could try to go further comparing $a_n$ and $a_m$ in general.
