Base Case: Show that the statement holds true for n = 1. Verify if (3^(2*1) - 1) is divisible by 2^(1+2) but not by 2^(1+3).

Inductive Hypothesis: Assume that the statement is true for some arbitrary integer k >= 1. That is, assume (3^(2k) - 1) is divisible by 2^(k+2) but not by 2^(k+3). This means (3^(2k) - 1) = 2^(k+2) * m, where m is an odd integer.

Inductive Step: Prove that the statement holds true for n = k+1. Show that (3^(2(k+1)) - 1) is divisible by 2^((k+1)+2) but not by 2^((k+1)+3).

Manipulation: Rewrite (3^(2(k+1)) - 1) as (3^(2k+2) - 1) = 9 * 3^(2k) - 1 = 9 * (3^(2k) - 1) + 8 = 9 * (2^(k+2) * m) + 8 = 2^(k+2) * (9m) + 2^3 = 2^3 * (2^(k-1) * 9m + 1).

Conclusion: From the manipulation step, since m is odd, (2^(k-1) * 9m + 1) is odd. Therefore, 3^(2(k+1)) - 1 is divisible by 2^3, but not by 2^4, which confirms the statement for n = k+1.