The extension $k(x)$ over $k(x^p)$ has degree $p$

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Let $x$ be transcendental over $k$. I wonder how to show:

The extension $k(x)$ over $k(x^p)$ has degree $p$.

I know $y^p-x^p$ is a polynomial over $k(x^p)$ that has a root $y=x$, but I don't know how to prove that $y^p-x^p$ is irreducible/minimal over $k(x^p)$.

Let $y^p-x^p=f(y)g(y)$, we need to show that one of them is trivial. But things get really messy from here and I don't see a slick proof