Let F(k)= $2^{p^k}+1$ where p is prime and k in a natural number. Let b be the highest power of p that divides F(k). So, $F(k)=p^bq$ where q is a natural number and (p,q)=1. $2^{p^k}=p^bq-1$ Then,F(k+1)=$ 2^{p^{k+1}}+1 = ({2^{p^k}})^p +1 = (p^bq-1)^p+1 = p^{b+1}(q-pC_2p^{b-1}q^2+...+p^{(bp-b-1)}q) $ which is divisible by $ p^{b+1}$ if b≥1, but is never divisible by $ p^{b+2}$ as (p,q)=1.
$Now,\ if\ p|({2^{p^r}}+1)\ ,\ i.e.,\ p|((1+1)^{p^r}+1),\ i.e.,\ p|(3+∑p^rC_r)\ where\ 1≤r So, p|3 as each of $p^rC_r$ is divisible by p, which implies p=3.
Now F(0)=3, F(1)=$2^3+1=9\ is\ divisible\ by\ 3^1,\ F(2)=2^{3^2}+1= 513\ is \ divisible\ by\ 3^2 .$ So, by mathematical induction, F(k)= $2^{3^k}+1$ is divisible by $3^k$