Here is a second, more algebraic proof.
Consider the morphism $f:\mathbb A^1_k \to \mathbb A^n_k :t\mapsto (f_1(t),\ldots,f_n(t))$.
It corresponds to the morphism of $k$-algebras $\phi: k[T_1,...,T_n] \to k[T]:T_i\mapsto f_i(T) $.
There are now two cases:
a) If all the polynomials $f_i(T)$ are constant the image of $f$ is a point in $\mathbb A^n_k$, and so obviously a closed set.
b) If some $f_i$ is not constant, then $T$ is integral over $k[f_i(T)]$ and a fortiori over $k[f_1(T),...,f_n(T)]$.
In other words the morphism $\phi: k[T_1,...,T_n] \to k[T] $ is integral (and even finite).
So the morphism $f:\mathbb A^1_k \to \mathbb A^n_k $ is integral and thus closed. In particular $f(\mathbb A^1_k)\subset \mathbb A^n_k$ is closed
Despite appearances this proof is more elementary than the preceding one. It only uses that integral morphisms are closed, which follows from lying-over.
The simplicity of the first proof is a bit deceptive: it uses as a black box that projective space is complete.
This is often thought trivial because it corresponds to compactness over $\mathbb C $, but the algebraic proof of completeness is not trivial, nor, come to think of it, is the assertion that completeness is equivalent to compactness in the classical case ( a GAGA-type of assertion)