Determinant: Alternative Definitions Reference
Foundation for: Determinant: Continuity
Problem
Given a vector space $V$.
Consider an endomorphism $T:V\to V$.
The rank of an endomorphism:
$$\mathrm{rank}T:=\dim\left(\mathrm{im}T\right)$$
The determinant of an endomorphism:
$$\det T:=\text{???}$$
What would be a nice definition not relying on representations by matrices?
(I assume basic knowledge of Differential Geometry and Functional Analysis.)

Let $V$ be an $n$-dimensional vector space over the field $\mathbb{F}$. 
Given a linear map $T : V \to V$, there is an induced linear map $\bigwedge^nT : \bigwedge^n V \to \bigwedge^n V$ given by $\left(\bigwedge^nT\right)(v_1\wedge\dots\wedge v_n) = (Tv_1)\wedge\dots\wedge(Tv_n)$. As $\bigwedge^nV$ is one-dimensional, $\bigwedge^nT = k\operatorname{id}_{\bigwedge^nV}$ for some scalar $k \in \mathbb{F}$. This scalar is precisely $\det T$.

Let me summarise some facts about the vector spaces $\bigwedge^pV$ (see the Wikipedia article on exterior algebras for more information).
Given a vector space $V$ of dimension $n$, there is an associated vector space $\bigwedge^pV$ for any $0 \leq p \leq n$ called the $p^{\text{th}}$ exterior power of $V$. The elements of $\bigwedge^pV$ are linear combinations of terms of the form $v_1\wedge\dots\wedge v_p$ where $v_1, \dots, v_p \in V$. The symbol $\wedge$ is called the wedge product, and it satisfies skew-symmetry, i.e. $v_i\wedge v_j = -v_j\wedge v_i$. If $\{v_1, \dots, v_n\}$ is a basis for $V$, then $\{v_{i_1}\wedge\dots\wedge v_{i_p} \mid i_1 < \dots < i_p\}$ is a basis for $\bigwedge^pV$ and therefore the dimension of $\bigwedge^pV$ is ${n \choose p}$.
