Step 1: Recall when a square system has a unique solution.
For a square system \(Ax=b\) with \(A\) of size \(n\times n\), the system has a unique solution if and only if \(A\) is invertible, that is, \(\det(A)\neq0\) and \(\text{rank}(A)=n\).
Step 2: Test statement (A).
Since a unique solution is given, \(A\) must be invertible, so \(A^{-1}\) exists directly by this fact. Statement (A) is correct.
Step 3: Test statement (B).
If \(A\) is invertible, then \(A^m\) is also invertible for every positive integer \(m\), because a product of invertible matrices is invertible:
\[ (A^m)^{-1}=(A^{-1})^m \]
Since \(A^m\) is invertible, the system \(A^mx=b\) has the unique solution
\[ x=(A^{-1})^mb \]
Statement (B) is correct.
Step 4: Test statement (C).
Since \(A\) is invertible, \(\text{rank}(A)=n\). Since \(A^m\) is also invertible for every \(m\), \(\text{rank}(A^m)=n\) as well. So
\[ \text{rank}(A)=\text{rank}(A^m)=n \]
for every \(m=1,2,3,\ldots\). Statement (C) is correct.
Step 5: Test statement (D).
The augmented matrix \([A\,|\,b]\) has \(n\) rows and \(n+1\) columns, so its rank can be at most \(n\). Since \(A\) is invertible, \(\text{rank}(A)=n\) already, and adding one more column cannot raise the rank of an \(n\)-row matrix above \(n\). So
\[ \text{rank}([A\,|\,b])=n=\text{rank}(A) \]
The two ranks are equal, not one strictly less than the other. Statement (D) is incorrect.
Step 6: Final conclusion.
The correct statements are (A), (B), and (C).
\[
\boxed{\text{(A), (B), and (C) are correct}}
\]