$A\in M_2(\mathbb C)$ and $A $ is nilpotent then $A^2=0$. How to prove this? $A\in M_2(\mathbb C)$ and $A $ is nilpotent then $A^2=0$. How to prove this?
I am not getting enough hints to start.

If $A^{n}=0$ but $A^{n-1}\ne 0$, then there exists $x \ne 0$ such that $A^{n}x = 0$ but $A^{n-1}x \ne 0$. Then $\{ x, Ax,\cdots, A^{n-1}x \}$ is a linearly independent set. To see why, suppose that
$$
              \alpha_{0}x+\alpha_{1}Ax+\cdots+\alpha_{n-1}A^{n-1}x = 0.
$$
By applying $A^{n-1}$ to the above, we arrive at $\alpha_{0}A^{n-1}x=0$, which gives $\alpha_{0}=0$. Then apply $A^{n-2}$ to conclude that $\alpha_{1}=0$, and so on. Because we're working on a two-dimensional space, then $A^{2} = 0$ must hold in order to avoid a contradiction of dimension.
