A field finitely generated as a k-algebra is a finite extension of k

Let $t_1,\dots ,t_r$ be a transcendence basis of $K$ over $k$. By [L1] there exists a nonzero $s\in k[t_1,\dots ,t_r]$ such that $K$ is integral over $R:=k[t_1,\dots ,t_r][1\mathrm{/}s]$.

Assume $r>0$. Because $K$ is a field and integral over $R$, every nonzero element $a\in R$ is a unit of $R$: the inverse $a^{-1}\in K$ satisfies a monic equation over $R$, and multiplying by a large power of $a$ rewrites that equation as $a^{-1}=-(c_{n-1}+c_{n-2}a+\cdots +c_0{a}^{n-1})\in R\mathrm{.}$ Thus $R$ would be a field, contradicting [L2].

Therefore $r=0$, so the transcendence basis is empty and $K\mathrm{/}k$ is algebraic. Since $K$ is finitely generated as a $k$-algebra, choose algebra generators $a_1,\dots ,a_m$; they are algebraic over $k$, and [L3] makes $K=k(a_1,\dots ,a_m)$ finite over $k$.