Every cyclic subspace contains an eigenvector The question is :

Let $X$ be a non-null vector.Then there exists an eigenvector $Y$ of $A$ belonging to the span of $\{X, AX, A^{2} X, ... \}$.

I have tried to the best of my ability to solve it. But I don't find any right way to proceed.Please help me.
I want to add a solution of my own question which I find just now in a pdf. Here's this :

Let $k$ be the least positive integer such that $X, AX, A^{2} X, ... , A^{k} X$ are linearly dependent. Now let us consider a relation $\sum_{i=0}^{k} c_{i} A^{i} X = 0$. Then we must have $c_k \neq 0$. Now let us consider a polynomial $g(t) = \sum_{i=0}^{k} c_{i} t^{i}$. Let $\beta_{1}, \beta_{2}, ... , \beta_{k}$ be the roots of the polynomial $g(t)$. Then $g(t) = c_{k} \prod_{i=1}^{k} (t - \beta_{i})$.

Hence, $\sum_{i=0}^{k} c_{i} A^{i} = g(A) = c_{k} \prod_{i=1}^{k} (A - \beta_{i} I)$.Taking $Y = (\prod_{i=2}^{k} (A - \beta_{i} I)) X$ , it is easy to see that $Y \neq 0$ by the minimality of $k$ and $(A - \beta_{1} I) Y = 0$. Hence the result follows.
But at last it is not clear to me why is $Y$ in the span of $\{X, A X, A^{2} X, ... \}$? Please anyone suggest me what is the trick behind it.
Thank you in advance.

Why $Y\ne 0$:   Let $h(t)=\prod_{i=2}^k(t-\beta_i).$ Then $\deg (h)<k$ and $h(t)$ is not identically $0$ so by the minimality of $k$ we have $0\ne h(A)(X).$ But $h(A)(X)=Y.$
