One way of defining the determinant of a endomorphism of a vector space $\varphi:V \to V$ is by using the action of $End(V)$ on the underlying $\mathbb{Z}$-graded vector space of the exterior algebra $\bigwedge^{\cdot}\left(V\right)$.This action is given by taking the canonical action of $\varphi$ on the underlying graded vector space of the tensor algebra and restricting it to the subspace $\bigwedge^{\cdot}\left(V\right)$. If $n=dim(V)$, then $\bigwedge^{n}\left(V\right)$ is $1$-dimensional, and any choice of basis identifies this space with the ground field $K$ and the action of $\varphi$ with the determinant of $\varphi$ expressed in that basis.

Now let $V_\cdot$ be a $\mathbb{Z}_2$-graded vector space and $\varphi:V_\cdot \to V_\cdot$ be a (parity preserving) endomorphism. Does the Berezinian of $\varphi$ have a description anything like the lines above? It cannot work "as is", since if $V_\cdot$ is purely odd, then $\bigwedge^{\cdot}\left(V\right)$ can be identified with the symmetric algebra on $\Pi V_\cdot$, which is has no top power, as $\Pi V_\cdot$ is purely even, but is there any relation at all?