= Solution
The quotient $A$ is a field generated as a $k$-algebra by the images $\alpha_1,\ldots,\alpha_n$ of the variables. We prove the relevant form of <Zariski lemma>.
Choose a maximal algebraically independent subfamily, say $\alpha_1,\ldots,\alpha_r$, and put
$$
K=k(\alpha_1,\ldots,\alpha_r).
$$
Every remaining $\alpha_j$ is algebraic over $K$. After clearing the finitely many denominators in their monic equations, there is a nonzero
$$
h\in k[\alpha_1,\ldots,\alpha_r]
$$
such that all $\alpha_j$ with $j>r$ are integral over
$$
R=k[\alpha_1,\ldots,\alpha_r,h^{-1}].
$$
Because $A$ is a field containing $R$ and is generated by these integral elements, $A$ is integral over $R$. If a field is integral over a subring, that subring is itself a field: for $0\ne c\in R$, an integral equation for $c^{-1}$ can be multiplied by a power of $c$ to express $c^{-1}$ in $R$.
If $r>0$, uncountability of $k$ lets us choose $a\in k$ such that $\alpha_1-a$ does not divide $h$. Then $\alpha_1-a$ is not invertible in the localization $R$, contradicting that $R$ is a field. Hence $r=0$, so every generator $\alpha_i$, and therefore every element of the finite algebraic extension they generate, is algebraic over $k$. Thus
$$
\boxed{\ A/k\text{ is algebraic}.\ }
$$
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