Your argument breaks down where you say that $(A^2)(A^3) =0$ implies that either $A^2 = 0$ or $A^3 = 0$. In general, there is no zero-product rule in linear algebra; it's easy for $AB$ to equal $0$ even if $A$ and $B$ are both non-zero. In bigger spaces, your theorem would be false. For instance, suppose you make a $5\times 5$ matrix which is all zeros except on the diagonal just above the main diagonal; for instance, let $A_{12}=A_{23}=A_{34}=A_{45}=1$ and everything else zeros. Then $A^5 =0$ but $A^2$ is not zero.
That being said, you are working only in the set of $2\times 2$ matrices. That's a pretty special case and the statement is surely true for that case. I don't know how far along in your textbook you are...one approach would be to observe that the minimal polynomial of a $2\times 2$ matrix can't be any larger than degree $2$; since the minimal polynomial must be a factor of $x^5$, it can only be $x^1$ or $x^2$.
If you haven't covered characteristic polynomials and minimal polynomials yet, you could say that, whatever the matrix, there is a basis in which the matrix is upper triangular. In that basis, the diagonal entries of $A^5$ are just the various values of $a^5$, where $a$ is one of the diagonal entries. Since $A^5$ is $0$, the diagonal entries in this basis must be $0$. This leaves only one possible non-zero entry, in the northeast corner. So $A^2$ must equal $0$ . There are some easy generalizations to bigger spaces--e.g., if $A$ is $m \times m$ and $A^n = 0$, where $n$ is bigger than $m$. Something like that.
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