[Math] Is $x^y$ always irrational if $x$ is rational and $y$ is irrational

Question

Prove or disprove:

"If $x$ is a rational number, and $y$ is an irrational number then
$x^y$ is irrational"

I am stuck with this, these are my steps.

let $x=2$ and $y=\sqrt{2}$

$\implies$ $x^y = 2^{\sqrt{2}} $

now if $x^y = 2^{\sqrt{2}} $ is irrational then we are done. But if this is rational then we can say:

let $x=2^\sqrt{2}$ (since we assume its rational) and let $y=\sqrt{2}$

$\implies$ $x^y = 2^{\sqrt{2}^{\sqrt{2}}} $ $=2^2$

this shows that if $x$ is rational and $y$ is irrational then $x^y$ is rational.

But I know that this is not true.
Where did I go wrong in this?

Answer 1

Hint (for an easy proof/disproof): what if $x = 1$?