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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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Not only does everyone know that $\pi$ is irrational, but everyone also knows that $\pi$ is transcendental :-) –  Robin Chapman Aug 12 '10 at 20:43
@Robin Chapman: Ok. Agreed. Ivan Niven's proof is awesome for the irrationality of $\pi$. –  anonymous Aug 12 '10 at 20:45
Does anyone have a link to a proof of transcendentality for completeness? –  Casebash Aug 12 '10 at 21:49
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. –  J. M. Aug 12 '10 at 22:25
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1 Answer

up vote 22 down vote accepted

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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@Moron: Ok, then if thats the case then why have they proved the seperate case of $\pi^{2}$ being irrational at planet math. They could have used your argument. –  anonymous Aug 12 '10 at 20:54
@Chandru: No idea. Perhaps they have it there for historical reasons. –  Aryabhata Aug 12 '10 at 20:57
There's a (relatively) simple proof that just happens to work for $\pi^2$ (and from which the $\pi$ case is an immediate corollary). –  Robin Chapman Aug 12 '10 at 20:57
@Robin Chapman: Thank you. –  anonymous Aug 12 '10 at 20:58
1768 pi irrational, 1882 pi transcendental. proving irrationality is easier than proving transcendence. using the fact that pi is transcendantal seems to be overkill (in my interpretation of OP's question) –  yoyo Mar 27 '11 at 20:52
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