Denote by $\Psi : k^n\to k^n$ the map of affine spaces corresponding to $\Phi$, and without loss of generality assume $\Psi(0) = 0$. Putting $M = (x_1,\ldots,x_n)$ and $N = (y_1,\ldots,y_n)$, this means that $\Phi^{-1}(N) = M$, so $\Phi(M) = N$ since $\Phi$ is surjective. We then get an induced map $\Phi_\ast:M/M^2 \Phi_a:M^a/M^{a+1} \to N/N^2$N^{a}/N^{a+1}$ for any $a\geq 1$. Here both $M/M^2$ M^a/M^{a+1}$ and $N/N^2$ N^{a}/N^{a+1}$ are $n$-dimensional $k$-vector spaces of the same dimension, and $\Phi_\ast$ \Phi_a$ is clearly thus an isomorphism since it is clearly surjective. But now if $\Phi(f) = 0$ for some $f$, then $\Phi(f) \in N$ and hence $f\in M$. We then have Choose $\Phi_\ast(f) = 0$, a contradiction. Thus a$ so that $\Phi$ is injective.
Geometrically, saying f$ lies in $\Phi$ is surjective means M^a$ but not in $\Psi$ is M^{a+1}$ (such an immersion, and so its differential $a$ clearly exists: it is an isomorphism. But if the degree of the lowest degree homogeneous piece of $\Phi$ fails to be injective, f$). We then $\Psi$ maps $k^n$ onto a lower dimensional subvariety of have $k^n$, \Phi_a(f) = 0$ and so its differential isn't an isomorphism.
I feel obliged to add that we are somehow still using "dimension theory" here$f\notin M^{a+1}$, in the sense contradicting that the dimension of affine space is encoded in the dimension of its tangent space. However, knowledge of chains of primes etc. $\Phi_a$ is clearly overkillan isomorphism.
