I just read the following proof of the fact that a determinant polynomial is irreducible from here:
Here is a proof using a little bit of algebraic geometry (we assume k is algebraically closed). It can be seen that $V(det)$ is irreducible in $k^{n^2}$. It is for example the image of the morphism $M_n(k)\times M_n(k) \rightarrow V(det)$, $(P,Q)\mapsto PI_{n-1}Q$ and the source is irreducible.
Hence as $k[X_{ij}]$ is a UFD we have $I(V(det)) = \sqrt{(p)}$ where p is an irreducible polynomial. So $det = p^k$.
To show that $k=1$ we use the fact that the differential of the determinant at a matrice A is $D(det)_A(.) = Tr(Adj(A).)$ where Adj(A) is the adjugate matrix. For $A$ such that $rk(A)=n-1$ we have $rk(Adj(A))=1$ so $D(det)_A(.)\neq 0$ (e.g. what about nonzero nilpotent matrices) which forces $k=1$.
The argument from the last part that shows that for $det = p^k$ the $k$ equals $1$ I do not understand.
In the answer we choose a $n \times n$ matrix $A$ with rank $n-1$ and observe that $rk(Adj(A))=1$.
Well, why this implies $Tr(Adj(A).) \neq 0$ and why this forces $k$ to be $1$?