If $\det(\mathrm{adj}\,A) \ne 0$, then $\det(A) \ne 0$. I'm trying to understand the reason why $A$ is invertible only if $\mathrm{adj}\,A$ is invertible.
That's what I have right now: $A\, \mathrm{adj}\,A = |A|\cdot I$.
So if we take $\det$ of both sides we get: $|A\,\mathrm{adj}\,A| = ||A|\cdot I|$
and then: $|A| \cdot |\mathrm{adj}\,A| = |A|^n$
but now I'm stuck...
Appreciate your help.

For simplicity put $\,B:=adj\, A\,$ , so:
$$AB=|A|\cdot I\Longrightarrow |A||B|=|A|^n$$
We're done, since 
$$|B|=0\Longrightarrow |A|^n=0\Longrightarrow |A|=0$$
