A number is divisible by 3 if the sum of the digits is divisible by 3. How about a number of the type (2^{2n} – 1), where n is a positive integer? Can we show that it is always divisible by 3?

Hint
No need to use divisibility by 3 rule here,
but a particular factoring rule:
a^{n} – 1 = (a – 1)(a^{n-1} + a^{n-2} + … + a + 1).

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