Property of nilpotent square matrix, $2\times 2$ Let A be matrix $2\times 2$ and take integer $k \geq 2$. Show that if $A^k = 0$ then we must have that $A^2 = 0$
 A: Suppose that $A$ has an inverse matrix. Then, multiplying the both sides of
$$A^k=O$$
by $\left(A^{-1}\right)^k$ gives us $I=O$ where $I$ is the identity matrix, which is a contradiction.
Hence, if we set $A=\begin{pmatrix} a&b\\c&d\end{pmatrix}$, then we have $ad-bc=0$. So, by the Cayley–Hamilton theorem, we have
$$A^2=(a+d)A-(ad-bc)I=(a+d)A.\tag1$$
Hence, we have
$$A^3=(a+d)^2A,\ \ A^4=(a+d)^3A,\cdots, A^k=(a+d)^{k-1}A.$$
Since $A^k=O$, we have 
$$a+d=0\ \ \ \text{or}\ \ \ A=O.$$
From $(1)$, in either case, we have $A^2=O$.
A: Here is a laboured way of explaining a more general case (you have $k=2$). But it relies on more general theory.
Presumably it is a matrix over a field $\mathbb F$. Once a basis is chosen for a vector space $V$ of dimension $k$ over $\mathbb F$, a $k\times k$ matrix $M$ represents a linear map on $V=\mathbb F^k$.
Let $I_n$ be the image of $V$ under $M^n$ with $I_0=V$. Note that for $n\ge 1$we have $M^{n+1}V=M^n(MV)$. Because $MV\subset V$ we then have $I_{n+1}\subset I_n$.
Now suppose $\dim I_n=\dim I_{n+1}\gt 0$. Since the dimensions are equal, and $I_{n+1}\subset I_n$ we have $I_{n+1}=I_n$, and $MI_n=I_n\neq 0$. This implies that $M^rI_n=I_n\neq 0$ for all positive integers $r$, which contradicts the fact that $M$ is nilpotent.
We therefore have $\dim I_n \leq \max(\dim V-n,0)$ and in particular $\dim I_k=0$ whence $M^k=0$
