Basis for field extension of $\mathbb{F}_{p^n}$ over $\mathbb{F}_p$

Question

Let $p$ be a prime. Let $f_p : \mathbb{F}_{p^n} \rightarrow \mathbb{F}_{p^n}$ be the Frobenius map ($f(a) = a^p$). I want to find a characteristic polynomial of $f_p.$

I think $A = \{1, p, p^2, ..., p^{n-1}\}$ is a $\mathbb{F}_p-$basis of $\mathbb{F}_{p^n}$. Since $$\mbox{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) = <f_p> \simeq \mathbb{Z}/n\mathbb{Z},$$ $f_p$ fixes elements in $\mathbb{F}_p.$

So $f_p(1) = 1, f_p(p) = p, ..., f_p(p^{n-1}) = p^{n-1}$ (using $f(p) = p $ and $f$ is a linear map). So the characteristic polynomial is $(1-t)^n.$

But I saw here https://books.google.com/books?id=lrMOjbLhR2IC&pg=PA8&lpg=PA8&dq=diagonalizable+over+Fq+if+n+divides+p-1&source=bl&ots=jpC-9d3oWJ&sig=sc2D6D5igoLw5LRHTWWRf9qP7_E&hl=en&sa=X&ved=0ahUKEwjZ3pLjzbDbAhVSHjQIHYV8CcgQ6AEIKjAA#v=onepage&q=diagonalizable%20over%20Fq%20if%20n%20divides%20p-1&f=false (page 9)

that minimal polynomial is $t^n - 1.$ (I guess it is the characteristic polynomial as well)

So my attempt is wrong ?

The argument used in the book involves normal basis theorem which I have never seen/proved before. Is there a way to build a basis and calculate the characteristic polynomial directly ?

Besides direct computational, is there a way to find a characteristic polynomail ?

If the vector space $V$ is of dimensional $n$, and a polynomial $p(x)$ of degree $n$ satisfying

(1) $p(f_p) = 0$

(2) deg$(p) = n =$dim$V$

(3) $p$ or $-p$ is monic

Is it possible to conclude that $p$ is the characteristic polynomail ?