$K$ is algebraically closed implies $\mathbb A^n (K) \setminus V(F)$ is infinite. Let $F$ be a non-constant polynomial in $K[x_1,...,x_n]$ and $K$ an algebraically closed field.

Show then $\mathbb A^n  (K)\setminus V(F)$ is infinite for every $n\geq 1$.

Here $\mathbb A^n(K)$ is the $n$-dimensional affine space over $K$ and $V(F)$ is the zero set of $F$.
My attempt is as follows. If $n=1$ then we have $F$ a one variable polynomial so it has finitely many roots, or equivalently $V(F)$ is finite. Hence $\mathbb A^1  (K)\setminus V(F)$ would be infinite since $\mathbb A^1(K)=K$ is algebraically closed. But for higher $n$ I'm stuck.
Thanks in advance.
 A: Proceed by induction on $n$. The case $n = 1$ follows from elementary algebra. Now suppose that the result holds for a fixed $n \geq 1$ and consider $f \in k[X_1,...,X_{n + 1}]$. Let $d$ be the maximal exponent of $X_{n+1}$ in $f$. You can express $f(X_1,...,X_{n + 1})$ as a polynomial in $(k[X_1,...,X_n])[X_{n+1}]$, say $f(X_1,...,X_{n+1}) = h_0(X_1,...,X_n) + h_1(X_1,...,X_n)X_{n+1} + ... +  h_d(X_1,...,X_n)X_{n+1}^d$, where$h_0,...,h_d \in k[X_1,...,X_n]$. By the induction hypothesis, there is an infinite subset $S \subseteq k^n$ on which some $h_i$ is nonzero, so for a given $s = (s_1,...,s_n) \in S$, $f(s, X_{n+1}) = f(s_1,...,s_n,X_{n+1})$ has at most $d$ roots in $k$. Since $k$ is infinite (a finite field cannot be algebraically closed), it must be true that for each $s \in S$ there exists some $x_s \in k$ such that $f(s, x_s) \neq 0$. So $\{(s, x_s) : s \in S\} \subseteq \mathbb{A}^{n+1} \setminus V(f)$ is infinite, and $\mathbb{A}^{n+1} \setminus V(f)$ must then itself be infinite.
A: $V(f+1)$ is disjoint from $V(f)$ since any $(x_1, ... x_n)$ that is a root of $f$ would satisfy $(f+1)(x_1, ... x_n) = 1$. Therefore, it's sufficient to show that for any non-constant polynomial $g$, $V(g)$ is an infinite set. By choosing an irreducible component of $V(g)$, we can assume $V(g)$ is irreducible (and therefore, that $g$ is prime).
To do this we use dimension theory. The dimension of $V(g)$ is the dimension of $k[x_1, ... x_n] / (g)$ (Hartshorne I.1.7), so $dim(V(g)) = n-1$ by Krull's principal ideal theorem.
However, a finite set of closed points has dimension 0 since its coordinate ring can be written as a product of fields. Therefore, if $V(g)$ is finite then it has dimension 0. This would require $n = 1$, but the theorem is already known for $n=1$ so we're done.
