Induction produces a polynomial subalgebra over which the affine algebra is integral

By [L1], choose generators $x_1,\dots ,x_n$ of $A$ as a $k$-algebra. We prove the theorem by induction on $n$.

Inductive hypothesis: assume the statement for every $k$-algebra generated by at most $n-1$ elements.

Base case $n=0$: then $A$ is the image of $k$. Because $A\mathrm{\ne }0$ and $k$ is a field, the structure map $k\to A$ is injective and hence identifies $A$ with $k=k[]$. Thus $A$ is integral over the empty polynomial algebra.

If $x_1,\dots ,x_n$ are algebraically independent over $k$, then $A=k[x_1,\dots ,x_n]$ already has the required form.

Otherwise there is a nonzero polynomial relation among $x_1,\dots ,x_n$. If $k$ is infinite, apply [L3] to make such a relation monic in the last variable after a triangular change. If $k$ is finite, apply [L4] to make such a relation monic in the last variable after the exponent substitution. In either case we obtain new generators $y_1,\dots ,y_n$ of $A$ such that $A$ is integral over $B:=k[y_1,\dots ,y_{n-1}]$ by [L5].

The algebra $B$ is generated over $k$ by $n-1$ elements, so the induction hypothesis yields algebraically independent elements $z_1,\dots ,z_d\in B$ such that $B$ is integral over $k[z_1,\dots ,z_d]$. Then [L2] and step 2.3 show that $A$ is integral over $k[z_1,\dots ,z_d]$.

Step 2.2 handles the algebraically independent case, while steps 2.3 and 3.1 handle the dependent case. Therefore the theorem holds for $n$ generators, and hence for every finite-type $k$-algebra.