Proposition   Let $R\subseteq S\subseteq T$ be commutative rings. If $R$ is noetherian, and T finitely generated as an $R$ -algebra and as an $S$ -module, then $S$ is finitely generated as an $R$ -algebra.

Lemma (Zariski's lemma)   Let $(L:K)$ be a field extension and $a_1,\ldots,a_n\in L$ be such that $K(a_1,\ldots,a_n)=K[a_1,\ldots,a_n]$ . Then the elements $a_1,\ldots,a_n$ are algebraic over $K$ .

Proof. The case $n=1$ is clear. Now suppose $n>1$ and not all $a_i,1\leq i\leq n$ are algebraic over $K$ .
Wlog we may assume $a_1,\ldots,a_n$ are algebraically independent and each element $a_{r+1},\ldots,a_n$ is algebraic over $D:=K(a_1,\ldots,a_r)$ . Hence $K[a_1,\ldots,a_n]$ is a finite algebraic extension of $D$ and therefore is a finitely generated $D$ -module.
The above proposition applied to $K\subseteq D\subseteq K[a_1,\ldots,a_n]$ shows that $D$ is finitely generated as a $K$ -algebra, i.e $D=K[d_1,\ldots,d_n]$ .

Let $d_i=\frac{p_i(a_1,\ldots,a_n)}{q_i(a_1,\ldots,a_n)}$ , where $p_i,q_i\in K[x_1,\ldots,x_n]$ .
Now $a_1,\ldots,a_n$ are algebraically independent so that $K[a_1,\ldots,a_n]\cong K[x_1,\ldots,x_n]$ , which is a UFD.
Let $h$ be a prime divisor of $q_1\cdots q_r+1$ . Since $q$ is relatively prime to each of $q_i$ , the element ${q(a_1,\ldots,a_n)}^{-1} \in D$ cannot be in $K[d_1,\ldots,d_n]$ . We obtain a contradiction.