I tried now the following:

Let $x\in V$ such that $\psi^k(x) \ne 0$. We will show that the set $\{\psi^{k-n}(x),\psi^{k-(n-1)}(x),\dots,\psi^{k}(x)\}$ is linearly independent for $0\leq n\leq k$, i.e., that it holds it when $c_0\psi^{k-n}(x)+c_1 \psi^{k-(n-1)}(x)+\ldots +c_k\psi^{k}(x)=0$ then $c_0=\ldots c_k=0$.


Base case: For $n=0$ we have the set $\{\psi^k(x)\}$. A non-zero element set is linealry independent.

Inductive hypothesis: We suppose that it holds for $n=i$:
The set $\{\psi^{k-i}(x),\psi^{k-(i-1)}(x),\dots,\psi^{k}(x)\}$ is linearly independent, i.e., $$c_0\psi^{k-i}x+\ldots +c_k\psi^k(x)=0\Rightarrow c_0=\ldots =c_k=0$$

Inductive step: We want to show that it holds for $n=i+1$:
$$c_0\psi^{k-(i+1)}(x)+c_1 \psi^{k-i}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^{k}(x)=0 \\ \Rightarrow \psi (c_0\psi^{k-(i+1)}(x)+c_1 \psi^{k-i}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^{k}(x))=\psi (0) \\ \Rightarrow c_0\psi^{k-i}(x)+c_1 \psi^{k-(i-1)}(x)+\ldots +c_{k-1}\psi^{k}(x)+c_k\psi^{k+1}(x)=0 \\ \Rightarrow c_0\psi^{k-i}(x)+c_1 \psi^{k-(i-1)}(x)+\ldots +c_{k-1}\psi^{k}(x)=0$$ then by the inductive hypothesis we get that the set $\{\psi^{k-i}(x),\psi^{k-(i-1)}(x),\dots,\psi^{k}(x)\}$ is linearly independent, and so $c_0=\ldots c_{k-1}=0$

So, from $c_0\psi^{k-(i+1)}(x)+c_1 \psi^{k-i}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^{k}(x)=0$ we get that $c_k\psi^{k}(x)=0$ and since $\psi^{k}(x)\neq 0$, it follows that $c_k=0$.

Therefore, $c_0=c_1=\ldots =c_{k-1}=c_k=0$.

So, the set $\{\psi^{k-n}(x),\psi^{k-(n-1)}(x),\dots,\psi^{k}(x)\}$ is linearly independent for each $n$.

For $n=k$ we get that the set $\{x,\psi(x),\dots,\psi^{k}(x)\}$ is linearly independent.


Is this correct?