Basic field theory exercise:
Let $K$ be a field and $x \in K$ transcendental. Firstly I need to prove that $x^n$ is transcendental too. I proved that the map $ev_{x^n}:K[T] \longrightarrow K[x]$ is injective as a consequence of the injectivity of $ev_x$.
Now I need to show that $[K(x):K(x^n)]=n$
I'm still a bit confused by the definition of $K(x):=\{ \frac{f(x)}{g(x)}:f(T),g(T) \in K[T] \} \cong \{ \frac{f(T)}{g(T)}: f,g \in K[T] \}=K(T)$. I see that every element of $K(x):=\{ \frac{f(x^n)}{g(x^n)}:f(T),g(T) \in K[T] \}$ is "associated to an element of degree multiple of $n$ in K(T)".
I believe that basis of $K(x)$ over $K(x^n)$ looks like $B:=\left(\frac{f(x)}{g(x)}, x \cdot \frac{f(x)}{g(x)},x^2 \cdot \frac{f(x)}{g(x)},..., x^{n-1} \cdot \frac{f(x)}{g(x)} \right)$. Since $x^k=x^{k'} \iff k=k'$ linear independence should be clear. Showing that $<B>=K(x)$ looks more complicated but would anyway follow from $|B|=n$.
If I'm not mistaken the solution is pretty much there but I'm still quite confused so I would really appreciate any hint on how to make it more formal.
Recall that if $F/E$ is an extension such that $F = E(\alpha)$ for some $\alpha \in F$, then $F/E$ is a finite extension whenever $\alpha$ is algebraic over $E$; moreover, if $p(T) \in E[T]$ is the minimal polynomial of $\alpha$ over $E$, then $[F:E] = m := \deg p(T)$ and $1,\alpha,\dots,\alpha^{m-1}$ is a $E$-basis for $F$.
Now, notice that $K(x) = K(x^n)(x)$, and so it suffices to find an irreducible monic polynomial $p(T) \in K(x^n)[T]$ with $n$-degree such that $p(x) = 0$. In fact, you can consider $p(T) := T^n-x^n$, and once you have shown that $p(T)$ is irreducible, it follows that a $K(x^n)$-basis for $K(x)$ is given by $1,x,\dots,x^{n-1}$.