The first term of the sequence is given as

64 = 2^6From the formula,

2nd term = \(2^2 * 2^6\)

3rd term = \(2^3 * 2^2 * 2^6\)

4th term = \(2^4 * 2^3 * 2^2 * 2^6\)

There is a

pattern with 2^6 constant for every term and the power of 2 is raised consecutively from \(2^2 to 2^n\)

Hence, \(2^1^1 = 2^6 * 2^2 * 2^3 ...........*2^1^1\)

similarly \(2^8 = 2^6 * 2^2 * 2^3.........*2^8\)

when finding the ratio everything cancels leaving only \(2^9 * 2^10 * 2^1^1 = 2^3^0\)

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This is my response to the question and may be incorrect. Feel free to rectify any mistakes