Use induction.

Put , and

Then and so the base case holds.

Now suppose for some , then:

............

The first of the terms is divisible by by assumption, and as a odd power of two is congurent to modulo and its square is congruent to modulo the second factor is divisible by and so we have is divisible by .

Which allows us to conclude by mathematical induction that is divisible by for any

.