Show that $2^{2^n}+5$ is divisible by $3$ for all $n \geq 1.$ This is part of a larger proof that I am trying to do, but this is a fact I'm trying to wrap my head around and prove. Based on a Sage program, I was able to deduce by pattern that $2^{2^n}+5$ is divisible by $3$ for all $n \geq 1.$ If you factor numbers of this form until $n=6$ on Sage, you get that a prime factor of all of them is $3.$ I tried proving this through induction, but I couldn't quite figure out how to make the argument for the inductive step. I know modular arithmetic is the way to go, but it isn't something we've studied yet in class! Thank you for your help.

$2^{2^n}+5 \equiv (-1)^{2^n}-1 \mod 3$.
If $n\ge1$, the exponent is even, and the expression becomes $1-1\equiv 0 \mod 3$.
