Show that every function $f\ \in \mathcal{F}(F,F)$ is uniquely a polynomial of degree $\leq q−1$ - (Approach) Let $F$ be a finite field with $q$ elements. Show that every function $f \in \mathcal{F}(F,F)$ is uniquely a polynomial of degree $\leq q−1$ with coefficients in $F$. That is, $\mathcal{F}(F,F)=Pol_{q−1}(F)$
This is a bonus question that I found on my school's website framed in terms of Linear Algebra. I've been working through Friedberg,Insel,Spence's - Linear Algebra for reference of the level that I'm at in terms of approaching this problem.
I did a search online for a solution and everything I've found to date revolves around Group Theory, not what I had in mind at this point.
From the tools that I have at my disposal up to this point the one technique that jumps out to me is somehow using the Lagrange Interpolation Formula, from the example and the exercises I've done involving it though it seems to be limited to getting a unique representation of a polynomial function. But what this question is asking is a representation for any function, so that is beyond just polynomials. Is there a way to use the Lagrange Formula or am I stuck until I learn some more tools?
 A: One way of approaching this question is to analyse the natural map $\phi:F[x]\rightarrow \mathcal{F}(F,F)$ which interprets a polynomial as a function from $F$ to $F$.
We want to say that this map, when restricted to polynomials of degree $\leq q-1$ is a bijection. For this, lets note that if $\phi(p(x))=\phi(q(x))$, then we have that $\phi(p(x)-q(x))$ is the zero function. So if we have that $p(x)-q(x)$ is nonzero, and yields the zero function once interpreted, then its a polynomial that has each $\alpha\in F$ as a root, so has at least degree $q$, since we have at least $q$ distinct roots.
This observation tells us that if $p(x)$, $q(x)$, are two distinct polynomials of degree $\leq q-1$, then $\phi(p(x))\neq \phi(q(x))$, so our map $\phi$ is injective when we restrict to this subspace. But now we can count, there are $q^q$ polynomials with coefficients in $F$ of degree $\leq q-1$, and there are $q^q$ functions from $F$ to $F$. So we have an injective map between finite sets of the same cardinality, so it must be surjective, and therefore bijective, giving the desired result.
