I want to show that $\gcd(F_n,n)=1$, where $F_n=2^{2^n}+1$. How to prove this?

I can show that that $\gcd(F_n, F_m)=1$ for any natural $n$ and $m$, and that $F_{n+1}=(F_n)^2-2F_n+2=F_0\dots F_{n-1}+2$, but I can't see how I can apply this to my problem. What am I missing?