Prove that the integers $n$ and $n, 2^{2^n} + 1$ are relatively prime. I need help to prove that
$$\gcd(n, 2^{2^n} + 1)=1,\ n = 1,2,\dots$$
I have no idea how start the proof.

First note the following: For any prime $p$ dividing $2^{2^n}+1$:
$$p|(2^{2^n}+1) \implies 2^{2^n} \equiv_p -1$$
$$\implies \left(2^{2^n}\right)^2 = 2^{2^{n+1}} \equiv_p 1$$
$$2^{2^n} \equiv_p -1; \ 2^{2^{n+1}} \equiv_p 1 \implies 2^{n+1} = Kc,$$
for some odd integer $K$, where $c$ is the smallest positive integer s.t. $2^c \equiv_p 1$ [and so $c$ divides $|(\mathbb{Z}/p\mathbb{Z})^{\times}|$ $= p-1$]. [To see why $K$ must be odd, as the equation $2^{2^n} \not \equiv_p 1$ holds, it follows that $2^n = \frac{2^{n+1}}{2}$ is not an integral multiple of $c$. So as $2^{n+1}=Kc$ with $K$ integral, it follows that $2^{n}= \frac{Kc}{2}$  with $\frac{K}{2}$ is not integral even though $K$ is integral, which gives $K$ odd.]
So as $c$ divides $p-1$, it follows that $Kc$ must divide $K(p-1)$, and so, as $2^{n+1}$ divides $Kc$, it follows that $2^{n+1}$ must divide $K(p-1)$ as well. Then, as $K$ is odd and $(K,2^{n+1})=1$, it follows that $2^{n+1}$ must not only divide $K(p-1)$ but $2^{n+1}$ must also divide $p-1$ itself. This gives $p > 2^{n+1}>n$. So any prime $p$ dividing $2^{2^{n}}+1$ must satisfy $p >2^{n+1}>n$. Thus,
there is no prime $p'$ dividing both $2^{2^n}+1$ and $n$, because all primes dividing $2^{2^n}+1$ are larger than $n$ and thus cannot divide $n$. This gives $(n,2^{2^n}+1)=1$.
