For every natural number $m$, $2^{2m}-1$ is divisible by $3$ Can anyone tell me if I missed any small details in this proof?
(basis) $2^{2(1)}-1 = 3*1$ which is divisible by 3. 
(induction) Fix $m>1$ so $p<m$. Hence for every natural number $p,$ $2^{2p}-1=3p.$ then for every natural number $(p+1),$ $2^{2(p+1)}-1=3(p+1)$ then
$2^{2(p+1)}-1=3(p+1)$
$2*2^{2p}-1=3(p+1)$
$2*2^{2p}-1=3(p+1)$
$2(3p+1)-1=3(p+1)$ by induction hypothesis
$2(3p+1)+2^2-3=3(p+1)$
$6p+3=3(p+1)$
$3(p+1)=3(p+1)$
Thus, the statement is true for the natural number n whenever it is true for any natural number less than n.

An option:
$m \ge 1.$
$2^{2m}-1 = (2^2)^m -1=$
$4^m -1= (1+3)^m-1.$
In the binomial expansion of $(1+3)^m$ all terms except the first term have a factor $3$, i.e.
$(1+3)^m=1^m +3k$, where $k \in \mathbb{Z^+}$.
Hence
$2^{2m}-1 = (1^m +3k) -1=3k .$
