A problem on rank of $A$ and $\operatorname{adj}(A)$ 
Let $A \in M_n(\mathbf{C})$ and let $B = \operatorname{adj}(A)$ be the adjugate matrix.
a) Prove that if $A$ is invertible, then so is $B$.
  b) Prove that if $A$ has rank $n-1$, then $B$ has rank $1$.
  c) Prove that if $A$ has rank at most $n-2$, then $B = O_n$.

a) We have that $A$ is invertible, so $\det(A) \neq 0$. The rank of $A$ is $n$. Also
$$
    A \operatorname{adj}(A)
  = \operatorname{adj}(A) A
  = \det(A) I.
$$
But what next? Please help me to solve the problem using elementary theory  of matrix algebra (also please do not use the rank–nullity theorem). 
EDIT:
b) Clearly $\det(A) = 0$. Then
$$
    A \operatorname{adj}(A)
  =  \operatorname{adj}(A) A
  =  \det(A) I
  =  0,
$$
i.e. $A \operatorname{adj}(A) = 0$. But how does this imply that $\operatorname{adj}(A)$ has rank $1$?

May I request for any alternative solution of the problem? Can we conclude the result from $A \operatorname{adj}(A) = 0$?

 A: 1) We know that $A\cdot \textrm{adj}(A) = \det A \cdot I_n$. That is, the inverse of $\textrm{adj}(A)$ is $\frac{1}{\det(A)} A$, hence $\textrm{adj}(A)$ is invertible.
2) Let A have rank $n-1$.
By Schur diagonalization theorem, $A=UTU^{-1}$, where $T$ is an upper triangular matrix and $U$ is a unitary matrix, by Schur's lemma.
Also, by Schur's lemma, $\textrm{rank}(T) = \textrm{rank}(A)$.
Since $T$ is a triangular matrix the rows of $T$ are all linearly independent of each other, hence one of the rows consists entirely of zeros because the rank of $T$ is $n-1$. Let this row be row $I$, where $1 \leq I  \leq n$.
Note that $\textrm{adj} A = \textrm{adj}(UTU^{-1}) = \textrm{adj}(U^{-1})\textrm{adj}(T)\textrm{adj}(U)$, but then $\textrm{adj}(U)$ and $\textrm{adj}(U^{-1})$ are of full rank, hence $\textrm{rank}(\textrm{adj} A) = \textrm{rank}(\textrm{adj}(T))$.
Now, imagine that we are calculating the cofactors of elements of row $j$. Suppose that $j \neq I$. Then, omitting the row $j$ and whichever column the element is on, will leave in the cofactor, the elements of row $I$, which are all zero. Since the cofactor is a determinant containing a zero row, it must have the value zero. Thus, row $j$ of $\textrm{adj}(T)$ is zero when $j \neq I$.
Suppose that $j=I$. Then, omitting the row $I$ and whichever column, we obtain some cofactor matrix $C$. Note that $C$ is of rank $n-1$, hence at least one of the cofactor elements is non-zero. That is to say, the $I$th row of $T$ is non-zero.
Thus, $\textrm{adj}(T)$ is identically zero except for one non-zero row. Hence $\textrm{rank}(\textrm{adj}(T)) = 1 = \textrm{rank}(\textrm{adj}(A))$ 
3)I'll go more loosely over this, if the logic has been understood:
If A has rank $2$, then on triangularizing you will get $2$ rows containing zero. Then the cofactor matrix will be empty, because every cofactor will contain at least one row of zeros. So $\textrm{adj}(A) = [0]_n$.
A: For a), of course we observe that $Adj(A)/\det(A)$ must be $A^{-1}$.
For b): note that $A$ has rank $n-1$. Conclude that we must have a null space with dimension at most $1$ (either apply the rank nullity theorem, or simply note that row-reducing to solve $Ax=0$ leaves only one free variable). However, since $A \; Adj(A)=0$, every column of the adjoint must be in the null space of $A$. Conclude that the rank of $Adj(A)$ is at most $1$.
Now, how do we conclude that $Adj(A)$ is not zero? For this, we need a little fact:

A matrix $A$ has rank less than $k$ if and only if every $k\times k$ submatrix has determinant zero

And with $k=n-1$, we see that not every entry of the adjoint can be zero.
For 3): directly apply the above fact.
