Let $V$ be a vector space over a field $F$ of characteristic $0$. A linear operator $T$ on $V$ induces a linear operator $\Lambda^k T:\Lambda^k V\to \Lambda^k V$ such that $\Lambda^k T(v_1\wedge \cdots\wedge v_k)=Tv_1\wedge\cdots\wedge Tv_k$ for all $v_1, \ldots, v_k\in V$.
If $n=\dim V$, then since $\dim(\Lambda^n V)=1$, we know that there is a unique $c\in F$ such that $\Lambda^n T(v_1 \wedge \cdots\wedge v_n)=c\cdot(v_1\wedge \cdots\wedge v_n)$. We call this constant the determinant of $T$.
From this definition of the determinant, it immediately follows that $\det(TS)=(\det T)(\det S)$ for all linear operators $S$ and $T$ on $V$.
Can we also easily show that $\det T^t=\det T$ for all $T\in \mathcal L(V)$?
Here $T^t$ denotes the transpose of $T$.