I'll assume that $A$ has order $n$ and is invertible, so that $|Adj(A)| = |A|^{n-1}\!$.
That's a consequence of $1/|A| = |Inv(A)| = |Adj(A)/|A|| = |Adj(A)|/|A|^n$, which is a corrected version of your initial thinking.

I was just looking at the n=3 picture till now. Got it now by seeing your answer.

As \(\displaystyle 1/|A| \) is scalar, it gets a power of n when Det is applied.
And as \(\displaystyle |Adj(A)|=|A|^{nâˆ’1}\!\), the result remains valid.