What is the relationship between $\ker(A), \ker(A^2), \ldots, \ker(A^n)$? If $\vec{y} \in \ker(A)$ then $A\vec{y} = \vec{0}$ so that $(A*A)\vec{y}= A*(A\vec{y})=\vec{0}$ and $\vec{y} \in \ker(A*A)$.
If $\vec{y} \in \ker(A*A*\cdots *A)$, then $(A*A*\cdots *A)\vec{y}=\vec{0}$ so that $(A*A*\cdots *A^{i-1})*(A\vec{y})=\vec{0}$ so that $\vec{y} \in \ker(A)$.
Does this mean that the $\ker(A)=\ker(A^2)=\cdots=\ker(A^n)$?
 A: Your first sentence is correct, and proves that $\ker A\subseteq \ker A^2$, and similar reasoning shows $\ker A^2\subseteq \ker A^3\subseteq\dots\subseteq \ker A^n$.
Your second sentence is not; what you said only allows you to conclude that $A\vec y\in \ker A^{i-1}$, not that $\vec y\in \ker A$.
To see an example where $\ker A\subseteq \ker A^2\subseteq \ker A^3$, but not the other way around, let $A(x,y,z)=(y,z,0)$. Then 
$$\ker A=\{(x,0,0):x\in\mathbb{R}\}$$
$$\ker A^2=\{(x,y,0):x,y\in\mathbb{R}\}$$
$$\ker A^3=\{(x,y,z):x,y,z\in\mathbb{R}\}=\mathbb{R}^3,$$
so the kernel grows each with each exponent of $A$.
A: No, this is not true. For a simple example, consider strict upper-triangular matrix $M$ whose entries are $1$s and $0$s. $M$'s kernel is usually not the whole space. But if you compute $M^{n}$, then you will find $M^{n}=0$ for large enough $n$. So these spaces are different. What you can assert is $$\ker(A^{i})\supset \ker(A^{i-1}),\forall i\ge 1$$
and I think the special case you suggested is equivalent to that $A$ is a projection after changing some coordinates.  
A: Your second argument is wrong: even though $A^{n-1}Ay=0$, it is not necessarily the case that $Ay=0$, because $Ay$ could be in $\operatorname{ker} A^{n-1}$ even if it is not $0$.
