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Everyone knows that $\pi$ is an irrational number, and one can refer to this page for the proof that $\pi^{2}$ is also irrational.

What about the highers powers of $\pi$, meaning is $\pi^{n}$ irrational for all $n \in \mathbb{N}$ or does there exists a $m \in \mathbb{N}$ when $\pi^{m}$ is rational.

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    The sketch of it at least: if $\pi$ were algebraic, Lindemann-Weierstrass would imply $\exp(2\pi i)$ is transcendental (proving e is transcendental is another story)... and you can fill in the rest.2010-08-12

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What Robin hinted at:

If $\pi^{n}$ was rational, then $\pi$ would not be transcendental, as it would be the root of $ax^{n}-b = 0$ for some integers $a,b$.

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    In my 2nd question, what I actually meant (but didn't think about until it was too late to edit my comment) is that can such a number $y$ can be explicitly exhibited, as opposed to simply proved to exist by some non-constructive method.2014-07-10