Here's an explanation for three dimensional space ($3 \times 3$ matrices). That's the space I live in, so it's the one in which my intuition works best :-).
Suppose we have a $3 \times 3$ matrix $\mathbf{M}$. Let's think about the mapping $\mathbf{y} = f(\mathbf{x}) = \mathbf{M}\mathbf{x}$. The matrix $\mathbf{M}$ is invertible iff this mapping is invertible. In that case, given $\mathbf{y}$, we can compute the corresponding $\mathbf{x}$ as $\mathbf{x} = \mathbf{M}^{-1}\mathbf{y}$.
Let $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ be 3D vectors that form the columns of $\mathbf{M}$. We know that $\det{\mathbf{M}} = \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$, which is the volume of the parallelipiped having $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ as its edges.
Now let's consider the effect of the mapping $f$ on the "basic cube" whose edges are the three axis vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$. You can check that $f(\mathbf{i}) = \mathbf{u}$, $f(\mathbf{j}) = \mathbf{v}$, and $f(\mathbf{k}) = \mathbf{w}$. So the mapping $f$ deforms (shears, scales) the basic cube, turning it into the parallelipiped with sides $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$.
Since the determinant of $\mathbf{M}$ gives the volume of this parallelipiped, it measures the "volume scaling" effect of the mapping $f$. In particular, if $\det{\mathbf{M}} = 0$, this means that the mapping $f$ squashes the basic cube into something flat, with zero volume, like a planar shape, or maybe even a line. A "squash-to-flat" deformation like this can't possibly be invertible because it's not one-to-one --- several points of the cube will get "squashed" onto the same point of the deformed shape. So, the mapping $f$ (or the matrix $\mathbf{M}$) is invertible if and only if it has no squash-to-flat effect, which is the case if and only if the determinant is non-zero.
A(adj A) = |A|(I)
|A(adj A)| = |(|A| I)|
|A| |adj A| = $|A|^n * |I|$
|A| |adj A| = $|A|^n $
case 1: if |A|$\neq0$
Then we get ,|adj A| = $|A|^{n-1} $
case 2: if |A|$=0$
Then,|adj A|$=0$
And, we again get |adj A| = $|A|^{n-1} $
Best Answer
In the matrix $A$, for two distinct indices $i,i'$ replace row $i'$ by row $i$ leaving all other rows the same, including row $i.$ Call the new matrix $B.$ Note that B has two idenical rows. Note that along row $i'$, the matrices $A$ and $B$ have the same co-factors. Expand $\det B$ along row $i'$. Since $B$ has two identical rows, its determinant is 0. The elements of $B$ along row $i'$ are the elements of $A$ along row $i$ and the co-factors of $B$ along row $i'$ are the co-factors of $A$ along row $i'$.