A Nice Exponential Problem | Japan Math Olympiad

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2²⁰⁴⁸ = 2^(2¹¹) = 2^(2ⁿ2¹¹⁻ⁿ)
2²⁰⁴⁸ = [2^(2ⁿ)]^(2¹¹⁻ⁿ)
2ⁿ = 11 - n => n = 3 (by inspection)
xˣ = 2²⁰⁴⁸ = (2⁸)^(2⁸)
*x = 2⁸ = 256*

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curiosity
xˣ = 2¹⁰⁴⁸⁵⁷⁶
1048576 = 2²⁰
2^(2²⁰) = 2^(2ⁿ2²⁰⁻ⁿ)
2^(2²⁰) = [2^(2ⁿ)]^(2²⁰⁻ⁿ)
2ⁿ = 20 - n => n = 4 (by inspection)
xˣ = 2¹⁰⁴⁸⁵⁷⁶ = 2^(2²⁰) = (2¹⁶)^(2¹⁶)
*x = 2¹⁶*