polynomials - Number of elements of a given trace in finite extension over prime subfield

Let $F = \mathbb{F}_{p^n}$ and $K = \mathbb{F}_p, p\in\mathbb{P}, n\in\mathbb{N}$.

Show that for every $k\in K$ there exist exactly $p^{n-1}$ elements in $F$ with a trace of $k$.

where the trace of an element $f\in F$ is:

$$\mbox{Tr}_{F/K}(f) = f + f^p + f^{p^2}+ \ldots +f^{p^{n-1}}$$
Trace is linear, which I have verified, therefore trace of $0$ is $0$ and also, assuming that the trace is zero for every $f\in F$, a polynomial of degree $p^{n-1}$ would have $p^n$ roots, so there must exist an element of nonzero trace and consequently, due to homogeneity of trace, for every $k\in K$ there exists an $f\in F$ with trace $k$.

Since $f^{p^n} = f$ we would get exactly $p^{n-1}$ elements of a given trace, but there is one problem. What guarantees that $f^{p^i}\neq f^{p^j}$ with $0 \leq i

Suppose for some $i

Unnecesary. $\mbox{Tr} : F\to K$ is onto, therefore $F/\mbox{ker Tr}\cong K$.

Answer

You already know Trace is a linear map and is not the zero map. Linear also implies Trace is an additive group homomorphism from a group of order $p^n$ to a group of order $p$. Being nontrivial, it has to be surjective, as the codomain order is a prime number. So the kernel has to be of order $p^{n-1}$. So the cosets of the kernel are all of the same cardinality, and so in each coset the trace is the same. Two different cosets cannot have the same trace as we need $p$ different trace values.