Subspace generated by an endomorphism In a proof for the Cayley-Hamilton theorem, I encountered the following linear subspace of some arbitrary finite dimensional vector space $V$ with an endomorphism $f:V \to V$ and fixed $v\in V$:
$$U_v := \text{span}(f^k(v): k\in \mathbb N)$$
Since $n:=\dim V < \infty$, we have $\dim U_v < \infty$ and hence already $l \in \mathbb N, l \leq n$ vectors of $U_v$ generate $U_v$. Now the (unproven) claim is that these vectors are
$$v_0 := v, v_1:=f(v), v_2:= f^2(v), \dots, v_{l-1} := f^{l-1}(v).$$
Now I don't fully understand this step. While of course it is obvious that we only need finitely many vectors generating $U_v$, why are we able to pick the first $l$ ones?  
I tried proving something like $$\text{span}(f^k(v):k\in \{0,...,l-1\}) = \text{span}(f^k(v):k\in I)$$
where $I\subseteq \mathbb N$ is some arbitrary index set with $| I | = l$.
But this is a really messy attempt and I don't really know if this goes into the right direction. Any help appreciated. 
 A: Let $v_i := f^i(v)$.

Lemma:
Assume $v_l \in \text{span}(\{v_0, ... v_{l-1}\})$.
Then, for every $\text{n} \in \mathbb{N}$, $v_{n} \in \text{span}(\{v_0, ... v_{l-1}\})$.
Proof:
Assume $v_k \in \text{span}(\{v_0, ... v_{l-1}\})$, so $v_k = \alpha_0 v_0 + ... +\alpha_{l-1} v_{l-1}$. 
Therefore, $v_{k+1} = f(v_{k}) = \alpha_0 f(v_0) + ... +\alpha_{l-1} f(v_{l-1}) = \alpha_0 v_1 + ... +\alpha_{l-1} v_{l} \in \text{span}(\{v_0, ... v_{l-1}\})$, because $v_{l} \in \text{span}(\{v_0, ... v_l\})$. 
Starting from $k = l$, the domino effect (or induction) completes the proof.

Let $l:=\text{dim}(\text{span}(f^k(v): k\in \mathbb N))$.
Claim: $$\text{span}(\{f^k(v): k\in \mathbb N\}) =  \text{span}(\{v_0, ... v_{l-1}\})$$
Proof:
If $\{v_0, ... v_{l-1}\}$ are independent, then the proof is complete (because these are $l$ independent vectors, $l$ is the dimension).
Otherwise, $\{v_0, ... v_{l-1}\}$ is a dependent set. 
Recall that using the lemma, it is enough to prove that $v_l \in \text{span}(\{v_0, ... v_{l-1}\})$.
$\{v_0, ... v_{l-1}\}$ is a dependent set, so one of the vectors there is a linear combination of the others. Assume $v_{l-1} \in \text{span}(\{v_0, ... v_{l-2}\})$. But then
$v_{l-1} = \alpha_0 v_0 + ... +\alpha_{l-2} v_{l-2}$
$ \rightarrow f(v_{l-1}) = \alpha_0 f(v_0) + ... +\alpha_{l-2} f(v_{l-2})$
$ \rightarrow v_{l} = \alpha_0 v_1 + ... +\alpha_{l-2} v_{l-1}$
and then $v_l \in \text{span}(\{v_1, ... v_{l-1}\}) \subseteq \text{span}(\{v_0, ... v_{l-1}\})$ and the proof is complete.
Otherwise, $v_{l-1} \notin \text{span}(\{v_0, ... v_{l-2}\})$. Still, $\{v_0, ... v_{l-2}\}$ is a dependent set. Assume $v_{l-2} \in \text{span}(\{v_0, ... v_{l-3}\})$. But this is impossible as 
$v_{l-2} \in \text{span}(\{v_0, ... v_{l-3}\}) \rightarrow v_{l-1} \in \text{span}(\{v_0, ... v_{l-2}\})$ in contradiction.
If follows that $v_{l-2} \notin \text{span}(\{v_0, ... v_{l-3}\})$.
The recurring argument gives that the set $\{v_0, ... v_{l-1}\}$ is independent, in contrary to the assumption that is is dependent. Therefore, 
$v_{l-1} \in \text{span}(\{v_0, ... v_{l-2}\})$, and so $v_{l} \in \text{span}(\{v_0, ... v_{l-1}\})$. As stated above, this completes the proof.
