The number \(\sqrt{2}^\sqrt{2}\), is it rational, an algebraic irrational or is it transcendental?

Recall that a rational number can be expressed as \(p/q\) for integers \(p, q\) with \(q \neq 0\). An algebraic irrational number \(\alpha\) is a solution to the equation \(P(x) = 0\), where \(P(x)\) is a polynomial with rational coefficients. A transcendental number is a real number that is irrational but not an algebraic irrational.

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