Using Fermats prime numbers to prove that there is infinitely many prime numbers A Fermat number $F_n$ is of the form $F_n = 2^{2^n} + 1$
Furthermore, $F_n = 2 + F_0F_1F_2......F_{n-1}$ Now I already proved that 
if $n \neq m$ then $\gcd(F_n,F_m) = 1$
Here is the proof
Without loss of generality , I assume that $m > n$ then we know that $$F_m = 2 + F_0F_1F_2........F_nF_{n+1}.....F_{m-1}$$ I assumed here that $n > 2 $ and $n < m-1$
Now we assume that $\gcd(F_n,F_m) = d$ then $d \mid F_n$ and $d \mid F_m$
Now if $d \mid F_n$ then $d \mid F_0F_1.....F_nF_{n+1}...F_{m-1}$ and so $d$ divides any linear combination of $F_m$ and  $F_0F_1.....F_nF_{n+1}...F_{m-1}$, In particular $d \mid Fm- F_0F_1.....F_nF_{n+1}...F_{m-1}=2$ and hence $$d \mid 2$$
Now since all Fermat numbers are odd then $d \neq 2$ and hence $d=1$
Now I want to use this to prove that there exists infinitely many primes.
I know that since Fermat numbers form an infinite sequence of increasing numbers and each Fermat number is relatively prime to all other Fermat numbers then we have infinitely distinct prime divisors for each composite Fermat numbers and we also add to those Fermat prime numbers.
I feel like there is a better mathematical argument that that novel that I wrote above :) 
 A: I think your proof is essentially fine. 
I'm not sure if you need to prove $F_n = 2+F_0F_1F_2\ldots F_{n-1}$, but I'll add that by induction. Base case:
$ F_0=3 \\
F_1 = 5 = 2+F_0$
Inductive step: Assume that the stated relation holds for $F_k$, so $F_k-2 = F_0F_1F_2\ldots F_{k-1}$. Then note that 
$$ \begin{align} 
(F_k)(F_k-2) &= (2^{2^k}+1)(2^{2^k}-1) \\
&= (2^{2^k})^2-1\\
&= 2^{2^{k+1}}-1\\
&= F_{k+1}-2\\
\end{align} $$
And so $F_{k+1} = 2+F_k(F_k-2) = 2+ F_0F_1F_2\ldots F_{k-1}F_k$ as required to complete the induction.
To show that for $m>n\ge 0, \gcd(F_m,F_n)=1$, your proof can be shortened, but doesn't change in essence:
$$\begin{align} F_m &= 2+F_0F_1F_2\ldots F_{m-1} \\
&=2+kF_n\\
\end{align}$$
Then for $d=\gcd(F_m,F_n)$ we have $d \mid F_m$ and $d \mid kF_n$, so also $d \mid 2$. We know that $F_m$ is odd $\Rightarrow d \ne 2$, giving $d=1$.
The infinite sequence of Fermat numbers therefore does not share any prime factors between terms, meaning there are an infinite number of primes.
