Proving $\deg(\det(A))=0$ given that there exists $B$ such that $AB=I$ where $A,B\in M_n(\mathbb F[x])$

Question

I noticed that the very similar question has been asked before, but has no answer to it, so I decided to ask anyway. I am instructed to prove this using $A\cdot\mathrm{Adj}(A)=(\det A)\cdot I$, but no matter how I tried to manipulate this, I don't know how to proceed. My progress so far is trivial.

Any help is appreciated.