TITLE: Polynomial map is surjective if it is injective
QUESTION [29 upvotes]: A friend of mine told me the following fact:

If $k$ is any algebraically closed field, then a polynomial map $f\colon k^n\to k^n$ of affine space $k^n$ is surjective if it is injective.

The proof he told me was actually a logic argument, which I didn't really understand, so I can't fully reproduce it here. The idea was that since it is the first order statement, it is enough to prove it for fields of characteristic $p>0$ for infinitely many $p$. Then by some reason it was enough to prove it for locally finite fields, and it somehow reduces to the same statement about finite fields $k$, where it is already obvious.
I am very far from logic, so I would like to see a nice algebraic (or geometric) proof of this fact. Do you know how to do that?
Thank you very much!

REPLY [6 votes]: For an argument that doesn't use model theory, see for instance paragraph 4.1 "Injective endomorphisms are surjective" in Arno van den Essen, Polynomial Mappings and the Jacobian Conjecture. Here the more general claim is proven that every injective endomorphism of an algebraic variety over an algebraically closed field is surjective.
It is still, as far as I can see, a similar argument to the model theoretic argument: injectivity and non-surjectivity can be described using polynomial constraints, which makes it possible to reduce this to the case of a finite field. But it is a purely algebraic argument.
If people are interested, I can try to reproduce the proof here; it will be shorter than the argument by Alex Youcis above, but still relatively lengthy.