Nilpotent matrix and basis for $F^n$

Layout: Suppose the set is linearly dependent.Then there exist scalars $\alpha _0, \ldots, \alpha _{n-1}$ not all null, such that $\alpha _0v+\ldots +\alpha _{n-1}A^{n-1}v=0_{n\times 1}$.

Since not all of them are null, you can take the smallest $j\in \{0,\ldots ,n-1\}$ such that $\alpha_j \neq 0$. (See spoiler below).

This transforms the equality above in something with less terms. Multiply it by an appropriate power of $A$ and find a contradiction.

Spoiler:

Since $\alpha _0= \ldots = \alpha _{j-1}=0$ the equality can be rewritten as $\alpha _jA^jv+\ldots +\alpha _{n-1}A^{n-1}v=0_{n\times 1}$. Now multiply this by $A^{n-j-1}$, (note that $n-j-1\ge 0$). Can you conclude?