Let dim $V = n$ and $T: V \rightarrow V $ be a linear map such that its only invariant subspaces are $0$ and $V$. Prove that {$v, T(v), ..., T^{n-1}(v)$} is a basis of $ V $, where $v \neq 0$.
I assumed that as this is a set of n vectors, it is sufficient to show that if they are either linearly independent or spanning, then it is a basis of V.
I tried to prove linear independence by contradiction. Assuming linear dependence,
$$ \exists j \in \{0,..,n-1\} $$ such that $$ \lambda_j T^j = \sum_{i = 0}^{j-1} \lambda_i T^i(v)+ \sum_{i = j+1}^{n-1} \lambda_i T^i(v) $$
$$ \Rightarrow T^j(v) \in span\{T_i(v): i\neq j , i\in\{0,...,n-1\} \}$$
But from here I didn't know where to use the fact that the only invariant subspaces are $0$ and $V$
You don't need Cayley-Hamilton. What you need to know is that the minimal polynomial of $T$ with respect to $v$ has degree $< n$. This follows since the set of $n + 1$ vectors: $\{v, Tv, \dots, T^nv\}$ is linearly dependent so you get some polynomial
$$ (a_0 + a_1T + \dots + a_mT^m)v = 0$$
for some $m \le n$. Now just multiply by $T^{n-m}$ so you can solve for $T^nv$. This shows that $\operatorname{span}\{v, Tv, \dots, T^{n-1}v\}$ is $T$-invariant.
Of course, since at the end we must have $\operatorname{span}\{v, Tv, \dots, T^{n-1}v\} = V$ it is necessary that $m = n$ but we don't need to know this to complete the proof.