Proof verification: Weak Nullstellensatz

I am trying to prove the following statement:

Let $$k$$ be an algebraically closed field, and let $$A = k[ X_1, \ldots , X_n]$$. If $$I$$ is an ideal of $$A$$ or $$A$$, define $$V(I) = \{ \mathbf{x}\in k^n | f( \mathbf{x} ) = 0 \hspace{5pt} \forall \hspace{5pt} f \in I\}$$.

Then $$V(I) = \emptyset \implies I = A$$

Attempt at proof:

Let $$k$$ be an algebraically closed field, and let $$A_n = k[ X_1, \ldots , X_n]$$. We proceed by induction on $$n$$. The base case of $$n=1$$ follows (with some effort) from $$k$$ being algebraically closed.

So assume $$n>1$$. Suppose for contradiction that there exists a proper ideal $$I$$ such that $$V(I)=\emptyset$$. Then we may assume $$I$$ maximal because all ideals are contained in maximal ideals and inclusion on ideals corresponds to reverse-inclusion on $$V$$.

Consider then $$\varphi: A_n \rightarrow A_{n-1}$$, such that $$X_i\mapsto X_i \hspace{5pt} \forall 1\leq i and $$X_n \mapsto 0$$, and $$\varphi$$ fixes $$k$$. This is a surjective ring homomorphism, so the image of $$I$$, call it $$I'$$, is a maximal ideal of $$A_{n-1}$$ or $$A_{n-1}$$ itself. At the same time, $$V(I') \hookrightarrow V(I)$$ by $$(x_1,x_2, ... , x_{n-1}) \mapsto (x_1,x_2, ... , x_{n-1},0)$$, so $$V(I)= \emptyset \implies V(I')=\emptyset$$. Now by our induction hypothesis we have that $$I'=A_{n-1}$$. Therefore $$1 \in I'$$, which in turn implies $$\exists f \in I$$ such that $$\varphi(f)=1$$, so $$f$$ must be a polynomial in $$X_n$$ only. If $$f$$ is constant then we are done. If not then by the maximality of $$I$$ and algebraic closedness of $$k$$, we have that $$X_n - c \in I$$ for some $$c \in k$$.

Note also that $$c \neq 0$$ because then $$\varphi(f)=0 \neq 1$$.

Now repeat the same argument in the paragraph above except letting $$\varphi(X_n)=c$$ instead. This yields that $$X_n-c' \in I$$ for some $$c' \neq c$$. $$I$$ is closed under addition so $$X_n-c - (X_n - c') = c'-c \in I$$. Hence $$I=(1)$$. Done.

So is this a valid proof? I suspect that it is too simple to be correct, but I cannot find any problem with it.

. . . $$\varphi(f)=1$$, so $$f$$ must be a polynomial in $$X_n$$ only.
From $$\varphi(f)=1$$, all you get is $$f=gX_n+1$$, for some $$g\in A_n$$.