Prove that $A^k = 0 $ iff $A^2 = 0$ Let $A$ be a $  2 \times  2 $ matrix and  a positive integer $k \geq 2$. Prove that $A^k = 0 $ iff $A^2 = 0$.
I can make it to do this exercise if I have $ \det (A^k) = (\det A)^k $. But this question comes before this.
Thank you very much for your help!

The solution using the minimal polynomial and Cayley-Hamilton is a bit of an over-kill and somewhat of a magic solution. I prefer the following non-magic solution (for the non-trivial implication, and there is no need to assume the matrix is $2\times 2$). 
Think of $A$ as a linear operator $A: V \to V$ with $V$ an $n$-dimensional vector space, and suppose $A^t=0$. We'll show that $A^n=0$. Now, for each $m\ge 1$ consider the space $K_m$, the kernel of $A^m$. It is immediate that $K_{m}\subseteq K_{m+1}$. Since $A^t=0$ it follows that $K_t=V$. It is also immediate that if $K_m=K_{m+1}$, then $K_m=K_{r}$ for all $r>m$. 
Thus, the sequence of kernels is an increasing sequence of subspaces, that stabilizes as soon as the next step is equal to the previous one, and it is eventually all of $V$. Now we will use the fact that $V$ is of dimension $n$. Considering the dimensions of the kernels, the above implies that the sequence of dimensions is strictly increasing until it stabilizes. Since it reaches $V$, the dimensions reach $n$ and since the dimension of $K_1$ is not zero, the sequence of dimensions starts at $1$ or more. That means it has to stabilize after no more than $n$ steps, and thus $K_t=V$ for some $t\le n$. But then $K_n=V$, and thus $A^n=0$.
