{$v,f(v),f^2(v),ldots,f^{n-1}(v)$} is a basis of $V$ if the minimal polynomial of $f$ is equal to the characteristic polynomial of $f$

Regarding your comment: I will remove the subscript $i$ to simplify notation.

At the relevant point in the proof, we are given that $T|_{V}$ has minimal polynomial $P^m$, where $P$ is irreducible, and we must show the existence of a cyclic generator of $V$, i.e., a vector $vin V$ such that the minimal degree monic polynomial $Q$ such that $Q[T](v)=0$ has $deg(Q)=dim(V)$.

Let $d = dim(V)$. Note in particular that the degree of this polynomial $Q$ is the dimension of the span of ${v,Tv,dots,T^{d-1}v}$.

As the minimal polynomial of $T|_{V}$ is $P^{m}$, this polynomial $Q$ divides $P^{m}$ for any vector$~vin V$: only powers of $P$ can occur. It therefore suffices to find a vector$~vin V$ for which $P^{m-1}[T](v)neq0$: if $P^{m-1}[T](v)neq0$, then we know that $Q$ divides $P^m$ but does not divide $P^{m-1}$, which means that we must have $Q = P^m$. However, that means that $deg(Q) = deg(P^m) = d$, which means that the set ${v,Tv,dots,T^{d-1}v}$ spans $V$.