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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?

Hyobin Lee
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    Do we even know whether or not $\pi^{\pi}$ is rational? – André Nicolas Jul 16 '16 at 07:04
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    I have a feeling that this a rather difficult question, but my understanding of transcendence is only superficial. – Jyrki Lahtonen Jul 16 '16 at 07:08
  • @AndréNicolas Surely any irrational exponent leads to an irrational outcome...but you make me question that assumption. – Jared Jul 16 '16 at 07:12
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    @Jared Not at all!!! $e^{\ln 2} = 2$. $e, \ln 2$ are both irrational. – Caleb Stanford Jul 16 '16 at 07:13
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    @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. – Caleb Stanford Jul 16 '16 at 07:14
  • @6005 Sure, but there are an infinite number of solutions to $e^{\ln(2)}$... – Jared Jul 16 '16 at 07:14
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    @Jared ???????????????????????? We're not talking in the complex numbers here. Natural log means natural log. – Caleb Stanford Jul 16 '16 at 07:15
  • @6005 OK, I'm almost certainly in the wrong here but I generally think of exponents in terms of complex numbers, sorry. – Jared Jul 16 '16 at 07:16
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    It's not known if $\pi^{\pi}$ is rational or not – ClassicStyle Jul 16 '16 at 07:28
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    @Jared Yes, you are wrong :) $\ln$ denotes the real natural log. Complex log must be explicitly specififed and usually is denoted with other notation. – Caleb Stanford Jul 16 '16 at 07:31
  • I suppose that $e^\pi\stackrel ?\in\Bbb Q$ is an open problem as well – Hagen von Eitzen Jul 16 '16 at 08:16
  • @HagenvonEitzen: actually, $e^\pi$ is transcendental by the Gelfond-Schneider theorem; see https://en.wikipedia.org/wiki/Gelfond%27s_constant – Dejan Govc Jul 16 '16 at 09:38
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    @HyobinLee: According to https://en.wikipedia.org/wiki/Yuri_Valentinovich_Nesterenko, transcendence of $e^\pi$ over $\mathbb Q(\pi)$ is also known (unless I'm missing something). – Dejan Govc Jul 16 '16 at 12:26
  • @DejanGovc Thank you! That answered my revised question! – Hyobin Lee Jul 17 '16 at 16:21

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