If k > 2, and n = 2^{k-3}, show that 3^{n} is not congruent to 1 (mod 2^{k}).
Old thread but if you're still interested: there are several ways to do this.
1) By induction for assume
etc.
2) A rather nice way is to consider
therefore the highest power of 2 dividing is 2
It follows that is the highest power of 2 dividing , so cannot divide