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Is $\pi^\pi$ algebraic over $\mathbb{Q}(\pi)$?

I have a feeling that it's a rather easy question, but since my understanding of field extensions is only superficial I really can't handle this original question.

*edit1. I read the comments and realized that I had better retreat and attack the transcendence of $e^\pi$ over $\mathbb{Q}(\pi)$ instead. In line with the first question, I have Taylor expansion in mind. I know that a Taylor expansion with coefficients in $\mathbb{Q}$ does not yield a rational output for a rational input in general. So that could be the key to a refutation. Any ideas?

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    $\begingroup$ Do we even know whether or not $\pi^{\pi}$ is rational? $\endgroup$ Jul 16, 2016 at 7:04
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    $\begingroup$ I have a feeling that this a rather difficult question, but my understanding of transcendence is only superficial. $\endgroup$ Jul 16, 2016 at 7:08
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    $\begingroup$ @Jared Not at all!!! $e^{\ln 2} = 2$. $e, \ln 2$ are both irrational. $\endgroup$
    – 6005
    Jul 16, 2016 at 7:13
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    $\begingroup$ @Jared Generally proving irrationality and transcendality are QUITE nontrivial even for numbers like $e^\pi$, $e$, $\pi$, $\sqrt{2}^\sqrt{2}$. Many things are not known; the things that are known like Baker's theorem are quite difficult to understand the proofs of. $\endgroup$
    – 6005
    Jul 16, 2016 at 7:14
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    $\begingroup$ @Jared ???????????????????????? We're not talking in the complex numbers here. Natural log means natural log. $\endgroup$
    – 6005
    Jul 16, 2016 at 7:15

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