Let $K$ be a field. Let $K(X)$ be the field of rational functions of $x$. And let $K(X^n)$ be a subfield. Show that $[K(X):K(X^n)]=n$.
In my understanding, we can treat $K(X)$ as a $K(X^n)$-vector space. So we have the basis $B=\{1, x, \cdots, x^{n-1}\} $ which has $n$ elements. Thus, the degree of of the field extension is $n$.
Now I want to show that $B$ is a basis. For the linear independence, I'm stuck on the part commented below.
For spanning, I'm a bit confused. Given an element $y$ in $K(x)$, we want to show $y= a_{n-1}x^{n-1}+\cdots + a_0$ for $a_i\in K(X^n)$. In this case, is the result clear using mod $x^n$? Or do I still need to show other things?
Thanks for any help!