jmc

Let's first try to find the cycle of units digits of $2^n$, starting with $n=1$: $2,4,8,6,2,4,8,6,\dots$ . The cycle of units digits of $2^n$ is 4 digits long: 2, 4, 8, 6. To find the units digit of $2^n$, for any positive integer $n$, we simply need to find the remainder, $R$, when $n$ is divided by 4 ($R=1$ corresponds to the units digit 2, $R=2$ corresponds to the units digit 4, etc.) Since $2^{1000}÷4=2^{998}$ without remainder, the units digit of $2^{2^{1000}}$ is 6. Therefore, the units digit of $F_n=2^{2^{1000}}+1$ is $6+1=\boxed{7}$.