For which $c$ is $\pi^c$ know to be irrational? 
For which $c$ is $\pi^c$ know to be irrational?

Clearly $\pi$ is irrational and it is known that $\pi^2$ is irrational. However I believe it is not known if $\pi^{1/\pi}$ is irrational. This made me wonder what the full range of values $c$ is for which $\pi^c$ is known to be irrational.
 A: Not only $\pi^2$, it is in general known that if $0<c\in\mathbb{Q}$ then $\pi^c$ is irrational. Let $c=\frac{p}{q}$ where $p,q$ are positive integers. If $\pi^{\frac{p}{q}}$ was rational then $\pi^p=(\pi^{\frac{p}{q}})^q$ would be rational as well. But this would imply that $\pi$ is a root of $x^p-\pi^p\in\mathbb{Q}[x]$ which would contradict that $\pi$ is trancendental. 
A: $\def\A{\mathbb A}$You can use the Lindemann-Weierstrass theorem to construct lots of such numbers:

If $α_j\in\A$ are distinct algebraic numbers, then the exponentials $e^{α_j}$ are linearly independent over $\A$. 

Now chose $α_2=0$ and $a\neq0$ in such a way that $α_1=a\ln\pi$ is algebraic. Then 1 and $e^{α_1}=\pi^a$ are independent over $\A$, i.e. $\pi^a$ is transcendental and thus irrational.
Note: Dunno whether $\ln\pi$ is algebraic or not, at least that theorem does not contradict it because we know $\pi\in\A$: Assuming $\ln\pi\in\A$ would just yield 1and $\pi$ were independent over $\A$, hence no contradiction. 
