Showing that $\varphi: k[x_1,...,x_n] \rightarrow Map(k^n,k)$ is injective Given a formal expression $f\in k[x_1,...,x_n]$ we can consider it as a map from $k^n$ to $k$ by evaluating the indeterminatns on elements in $k^n$.
I want to show association, which we denote $\varphi$, is injective.
We proceed by induction: Suppose $f\in \ker(\varphi)$. Since $k$ is algebraically closed, $k$ is infinite. Therefore $f$ has an infinite number of zeros. It should follow from this that $f$ must then be zero in $k[x_1,...,x_n]$ but I don't see why this is the case.
Bonus confusion: I am trying to find a counter example in the case the $k$ is not algebraically closed. If we take $k=\mathbb{F}_p$ then it seems $x^p-x$ is a counter example. Whenever we evaluate it on an element in $\mathbb{F}_p$ we get zero, but i am wondering why we don't also count this as being zero as a formal expression in $\mathbb{F}_p[x]$?
 A: Preliminary: I'm going to write polynomials with capital letter variables, and their induced maps with lower case variables to make them easier to distinguish.
About your first question: $f$ having an infinite number of zeros does not imply that $f=0$ for $n\geq2$. Take $X_1^2+X_2^2-1\in\mathbb C[X_1,X_2]$ as a counterexample. It is zero on the (real) unit circle, which contains an infinite number of elements, but the polynomial itself doesn't vanish. If you're going to do induction, actually do it:
$n=1$ :
This is where what you tried doing actually works. A single variable polynomial has at most as many zeros as its degree, which is finite. Except if it is the $0$ polynomial. So $f=0$, and thus $\ker\varphi$ is trivial, so $\varphi$ is injective.
$n+1$ :
Now let $f\in k[X_1,\dots,X_{n+1}]=k[X_1,\dots,X_n][X_{n+1}]$ such that $f\in\ker\varphi$. This means $f(x_1,\dots,x_{n+1})=0$ for all $x_1,\dots,x_{n+1}\in k$. consequently, if we consider $f$ as a single variable polynomial in $k[X_1,\dots,X_n][X_{n+1}]$ (the coefficients are polynomials in $k[X_1,\dots,X_n]$), it also has an infinite number of zeros. You can take it from here.
About your second question: The answer to why we don't consider $X^p-X=0$ requires an understanding of what a formal polynomial is supposed to do in the first place. Basically, if $R$ is a commutative ring we want a polynomial $f\in R[X]$ to be something into which we can substitute things. And I'm specifically saying things, not elements of $R$. Because there's more stuff we might want to substitute for $X$. For instance, think back to linear algebra, where the minimal polynomial of a matrix $M$ was defined as the normed polynomial $\mu$ with minimal degree such that $\mu(M)=0$. So we substitute a matrix for $X$, not an element of the underlying field. To be precise, we want there to be evaluation homomorphisms: If $S$ is a ring extension of $R$ and $s\in S$ an element which commutes with every element of $R$, then $\varphi_s:R[X]\to S,~f\mapsto f(s)$ is a ring homomorphism. This is what polynomials must be able to accomplish to be rightfully called polynomials. So back to your example $X^p-X\in\mathbb F_p[X]$, we need to be open to the possibility that we don't substitute an element of $\mathbb F_p$, but an element of any ring extension of $\mathbb F_p$. And while substituting any element of $\mathbb F_p$ would yield $0$, substituting elements of a ring extension might not! For instance, if $x\in\mathbb F_{p^2}\backslash\mathbb F_p$, then $x^p-x\neq0$, since the corresponding polynomial can have at most $p$ zeros due to its degree, which are already contained in $\mathbb F_p$.
Another way to look at it: You're only considering the homomorphism $k[X_1,\dots,X_n]\to\operatorname{Map}(k^n,k)$. But for any commutative ring extension $k\subset S$, there is also a homomorphism $k[X_1,\dots,X_n]\to\operatorname{Map}(S^n,S)$. And while the former might be non-injective for $k$ not algebraically closed, the latter might actually be injective. But only if you consider $X^p-X$ and $x^p-x$ different things, where $x^p-x\equiv 0$ doesn't imply $X^p-X\equiv 0$
