how is it possible for $a^n\ne a$? If an ideal $a$ is an additive subgroup of a ring $A$  satisfying $Aa\subset a$, then in particular, for any $a_1, a_2, \dots, a_n \in A$ and any $x\in a$ 
$$y_1=a_nx\in a$$
 so 
$$y_2=a_{n-1}y_1=a_{n-1}a_nx\in a$$
$$y_3=a_{n-2}y_2=a_{n-2}a_{n-1}a_nx\in a$$
$$\vdots$$
$$y_n=a_1y_{n-1}=a_1\dots a_nx \in a$$ 
but, taking each of the $a_i$ to be in $a$, and calling $x=a_0$ we get then that $a_0\dots a_n \in a$. So, if the powers $a^n$  of $a$ are defined as the ideals generated by all products $a_0\dots a_{n-1}$, then how is it possible for $a^n\ne a$? 

Your argument shows that $a^n\subseteq a$, but it does not show that $a\subseteq a^n$, which may in fact be false.  For instance take $A=\mathbb{Z}$ and $a=2\mathbb{Z}$.  Then an element of $a^2$ is a sum of products of two even integers.  Any such product is divisible by $4$, and so is any sum of such products, so every element of $a^2$ is divisible by $4$.  In particular, $2\in a$ but $2\not\in a^2$.  (With slightly more work, you can show that in fact $a^2=4\mathbb{Z}$.)
