In positive charcteristic $p$, we know that for every field element $x\in\mathbb{F}_{p}$ we get $x^p = x$.
Then I think (and I might be wrong, but I don't see how) monomials of the form $t^{p^i}\in\mathbb{F}_p[t]$ for arbitrary $i\in \mathbb{N}$ are all the same, since the functions $p_0(t)=t$, $p_1(t)=t^p$, $p_2(t)=t^{p^2}$ and so on are all actually the same functions. Not? I mean certainly they are equal on all elements, that is
$$p_i(x)= p_j(x)$$ for all $x\in\mathbb{F}_p$ and $i,j\in\mathbb{N}_0$. But in textbooks on finite function field and such, they are treated as if they where different. But this seems to contradict the (poinwise) definition of a function in terms of evaluation on elements.
So would be great to get some clarification here.
There are indeed two different notions of a polynomial over a field $K$: the first is that of formal expressions of the form $\sum_{i=0}^k a_iX^i$ where $a_i \in K$, and the second is that of functions $f : K \to K$ where there exist $a_i \in K$ such that for all $x \in K$ we have $$ f(x) = \sum_{i = 0}^k a_ix^k. $$ If we write $K[X]$ for the formal expressions, and $\mathrm{Pol}(K)$ for the functions, then there is a mapping $K[X] \to \mathrm{Pol}(K)$ given by interpreting the formal expression as a function. In fact, it is a homomorphism whose kernel is precisely those polynomial expressions which evaluate as the zero function.
Note that if $K$ is infinite, then the mapping $K[X] \to \mathrm{Pol}(K)$ is a bijection, and we tend to identify the two sets without too much worry.
When $K$ is finite, identifying the sets makes little sense: the set $\mathrm{Pol}(K)$ is a finite set -- in fact the set of all functions $K \to K$ -- that does not catch the structure we find interesting about polynomials over $K$. For a concrete example, $\mathrm{Pol}(\mathbb F_2)$ only has four elements, and they can be written (as polynomial expressions) as $0, 1, X, X - 1$. There are no polynomials of degree at least $2$ to form any splitting fields!