Let $V$ be a $K$-space with a basis $\{v_{1}, v_{2}, v_{3}, \dots\}$. Prove that there exists an product such that $x \cdot v_{i} = v_{i + 1}$ for $i \geq 1$ and $V$ get a structure of $K[X]$-module.
I can see that $x^nv_{i}=v_{i+n}$. The product
$$ \biggl(\sum_{i = 1}^{n} a_{i}x^{i}\biggr) \cdot \biggl(\sum_{i = 1}^{\infty} k_{i}v_{i}\biggr) = \sum_{i = 1}^{n}\sum_{j = 1}^{\infty} a_{i}k_{j}v_{i + j} $$
works?
I'd like some advice to this problem.
In your case the product is $$(\sum_{i=1}^n a_ix^i)\cdot (\sum_{j=1}^\infty k_j v_j)=\sum_{i=1}^n\sum_{j=1}^\infty a_i k_j v_{j+i}$$
In general this is the construction used to obtain for example the rational canonical form.
If $\phi $ is any $K$-linear transformation, we obtain a $K[x]$-module in this way $$(\sum_{i=1}^n a_ix^i)\cdot (\sum_{j=1}^\infty k_j v_j)=\sum_{i=1}^n\sum_{j=1}^\infty a_i k_j \phi^i(v_{j})$$