$y$ trancendental over $K$, then $y^n$ transcendental over $K$

Question

I have to prove the following, but I have no idea were to start.

Let $K$ be a field and $y$ transcendental over $K$. Then:

1. $y^n$ with $n\in \mathbb{N}$ is transcendental in $K$.
2. $K(y)$ is an algebraic extension of $K(y^n)$ with degree $[K(y):K(y^n)]=n$.

Answer 1

1. If $y^n$ is algebraic over $K$, then there exists a monic polynomial $f(x) \in K[x]$ such that $f(y^n)=0$. But then $g(x):=f(x^n)\in K[x]$ is such that $g(y)=0$, which contradicts the fact that $y$ is transcendental over $K$.

2. The polynomial $f(x):=x^n-y^n$ is irreducible over $ K(y^n)$, and $f(y)=0$. Thus, $y$ is algebraic over $ K(y^n)$ and $[K(y):K(y^n)]=\deg f=n$.

Answer 2

Here is the answers:

1. Suppose the contrary, that $y^n$ is not transcendental for some $n\in N$. Then, it must satisfy some algebraic equation, e.g. there is such k and such $a_0,...,a_k$ so that $a_0+a_1y^n+...+a_k(y^n)^k = a_0+a_1y^n + ...+ a_ky^{nk} =0$. Note that the latter equation is a polynomial of y being zero, i.e. a contradiction because y is transcendental Are you sure about the other thing?? I doubt this holds...