About a step in the proof about determinant of adjugate matrix I was trying to understand this answer which is about $\det(\operatorname{adj}(A))$.
I came across a step: |A adj(A)|=|(|A|I)|. The next step was: |A||adj(A)|=|A|^n * I. I didn't get how |(|A|I)|= |A|^n * I.

If I understand your post correctly, you are asking about determinant of the matrix $|A|\cdot I$.
In general, if you multiply a row of matrix by a constant $c$, determinant is multiplied by the same constant.
Multiplying the whole matrix by a constant $c$ is the same as multiplying each row of this matrix by $c$. So for any constant $c$ and any $n\times n$ matrix $B$ you have
$$|cB|=c^n|B|.$$
See also here: Prove $\det(kA)=k^n\det A$
This is just a special case where $B=I$ and $c=|A|$. So you get that
$$\det(|A|\cdot I)=|A|^n.$$
The answer you linked to writes this as $|A|^n\cdot|I|$, which is the same, since $|I|=1$.
