Proof.
Let $F_m$ and $F_n$ two Fermat numbers, and assume $m<n$ Let $d$ a positive common divisor of $F_n$ and $F_m$ that is $$ d \mid F_m,\qquad d\mid F_n. $$

If $d\mid F_m$ then $d\mid F_1F_2\cdots F_{n-1}$ since some factor must be $F_m$ itself. But $F_n-F_1F_2\cdots F_{n-1}=2$ and so $d \mid 2$ Since $d$ is odd, we must have $d=1$

Therefore, the greatest common divisor of any two Fermat numbers must be $1$

Q.E.D.