Prove $4^n-1$ is divisible by $3$, for all $n\in\Bbb N$? Prove $4^n-1$ is divisible by $3$, for all $n\in\Bbb N$?
I started by assuming there exists some $k\in\Bbb N$ s.t. $4^n-1=3k\iff \dfrac{4^n}3-\dfrac 13=k$, so for $k$ to be a natural number, $4^n\equiv 1\mod 3$ must be true, but this tells us no new information, and I don't know how to follow from there. Any help would be appreciated.

With only middle school tools:
$$4^n-1=4^n-1^n=(\underbrace{4-1}_{\textstyle 3})(4^{n-1}+4^{n-2}+\dots +4+1).$$
