Since $\psi$ is a linear mapping we have that $\psi (0)=0$.

$$\psi^k (c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x))=\psi^k (0) \\ \Rightarrow c_0\psi^k(x)+c_1\psi^{k+1} (x)+\ldots +c_{k-1}\psi^{2k-1}(x)+c_k\psi^{2k}(x)=0 \\ \Rightarrow c_0\psi^k(x)=0 \\ \Rightarrow c_0=0 \ \ \checkmark$$

Inductive hypothesis: We suppose that it holds that for $i\leq m$ : $$c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0 \Rightarrow c_i=0, \forall 1\leq i\leq m$$

Inductive step: We want to shw that it holds for $i=m+1$:

From the inductive hypothesis we get that $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0 \Rightarrow c_i=0, \forall 1\leq i\leq m$. So $c_{m+1}\psi^{m+1}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$.

Then we have the following:

$$\psi^{k-m-1} (c_{m+1}\psi^{m+1}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x))=\psi^{k-1} (0) \\ \Rightarrow c_{m+1}\psi^k (x)+\ldots +c_{k-1}\psi^{2k-m-2}(x)+c_k\psi^{2k-m-1}(x)=0 \\ \Rightarrow c_{m+1}\psi^k(x)=0 \\ \Rightarrow c_{m+1}=0$$

So, we have that when $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$ then $c_i=0, \forall i$, i.e., the set $\{x, \psi (x), \ldots , \psi^k\}$ ist linearly independent.

This is the only way to show that the set is inearly independent, right?