Let $K$ be a field and let $x_1, \ldots, x_n$ be indeterminates over $K$. Prove that $K$ is algebraically closed in $K(x_1, \ldots, x_n)$.
Definition: Let $F$ be a field extension of $K$. We say that $K$ is algebraically closed in $F$ if every element of $F$ that is algebraic over $K$ is already in $K$.
My attempt:
We need to show for any $u \in K(x_1,...,x_n)$ there is $p(x) \in K[x]$ such that $p(u)=0,$ where $x_1,...,x_n$ are indeterminates over $K$. Let $u \in K(x_1,...,x_n)$, then we have $u=\displaystyle \frac{f(x_1,...,x_n)}{g(x_1,...,x_n)}$ where $g(x_1,...,x_n) \neq 0$. We note that $f(x_1,...,x_n), g(x_1,...,x_n) \in K[x_1,...,x_n].$ I can not go along from here, I can not find $f(x) \in K[x]$ satisfy $p(u)=0$.
Thanks for any help or hints