Solution.
Since $A$ has rank 1, it has at least one nonzero row, $r$ and each of its rows is a multiple of $r$. Therefore,
there is a $1\times n$ row vector $X$ for which $A=X^{T}\cdot r$. Now, for any $p\times q$ and $q\times p$ matrices
$B$ and $C$, the trace of $B\cdot C$ equals the traces of $C\cdot B$. Therefore, we have
1=\text{trace of $A$}=\text{trace of } (r\cdot X^{T}) = r\cdot X^{T}.
(The last equality holds since $(r\cdot X^T)$ is a $1\times 1$ matrix.) Thus,
A^2 = (X^T\cdot r)(X^T\cdot r) = X^T\cdot (r\cdot X^T)\cdot r = X^T\cdot r = A.