For given $n\times n$ matrix $A$ singular matrix, prove that $\operatorname{rank}(\operatorname{adj}A) \leq 1$

Since $A$ is singular, $\mbox{rank}A\leq n-1$.

Case 1: $\mbox{rank}A\leq n-2$. Then $A$ contains no invertible submatrix of order $n-1$. So every minor of order $n-1$ is zero. What can you conclude about $\mbox{adj}(A)$?

Case 2: $\mbox{rank}A= n-1$. By rank-nullity, we get $\dim\ker A=1$. Now $A\cdot \mbox{adj}(A)=0$ means that the range of $\mbox{adj}(A)$ is contained in $\ker A$. So...

This is an old question, but I would like to add a solution that only uses properties of matrices i.e. no abstract algebra.
Sylvester's rank inequality states that for two matrices $X,Y \in \mathcal{M_n(\mathbb{C})}$ we have that $\operatorname{rank}(XY)\ge \operatorname{rank}X+\operatorname{rank} Y-n$.
Since $A$ is singular, we have $\operatorname{rank}A\le n-1$.
We have the equality $A\cdot \operatorname{adj}A=O_n$.
Hence, from Sylvester's rank inequality $0=\operatorname{rank}(A\cdot \operatorname{adj}A)\ge \operatorname{rank}A + \operatorname{rank}(\operatorname{adj}A)-n$.(1)
Case 1. $\operatorname{rank}A\le n-2=>\operatorname{adj}(A)=O_n$ as in Julien's solution.
Case 2. $\operatorname{rank}A=n-1$.
From (1) we have that $\operatorname{rank}(\operatorname{adj}A)\le 1$, so $\operatorname{rank}(\operatorname{adj}A)\in \{0,1\}$.
If $\operatorname{rank}(\operatorname{adj}A)=0$, then $\operatorname{adj}A=O_n$ and this contradicts $\operatorname{rank} A=n-1$.
Hence, $\operatorname{rank}(\operatorname{adj}A)=1$ in this case.
In both cases $\operatorname{rank}(\operatorname{adj}A)\le 1$.