I was recently considering this statement:
Let $V$ be a finite-dimensional $k$-vector space, and let $\phi:V\to V$ be an endomorphism. Suppose that $W\subseteq V$ is a subspace that is stable under $\phi$, i.e. such that $\phi(W)\subseteq W$.
Let $\psi:W\to W$ be the restriction of $\phi$ to $W$, and let $\rho:(V/W)\to(V/W)$ be the induced map on the quotient space $V/W$. Then $$\det(\phi)=\det(\psi)\det(\rho).$$
I came up with a proof, but it required choosing bases (ick!): if $\{w_1,\ldots,w_r\}$ is a basis for $W$, and $\{v_1,\ldots,v_s\}$ a set in $V$ that maps down to a basis of $V/W$, then their union $\{w_1,\ldots,w_r,v_1,\ldots,v_s\}$ is a basis for $V$. Expressing $\phi$ as a matrix in this basis, it is a block matrix of the form $$\begin{bmatrix} A & B\\ 0 & C \end{bmatrix}$$ because $\phi(W)\subseteq W$. But $A$ is the $r\times r$ matrix representing the action of $\psi$ on $W$, and $C$ is the $s\times s$ matrix representing the action of $\rho$ on $V/W$, and by properties of block matrices, we have $$\det(\phi)=\det\left(\begin{bmatrix} A & B\\ 0 & C \end{bmatrix}\right)=\det(A)\det(C)=\det(\psi)\det(\rho).$$
All well and good, but can someone tell me how to prove this statement the "right" way (via the exterior power functor, exact sequences, etc.?)
One way to see this is to interpret the determinant as the unique map between exterior powers which is normalized to send the identity map to $1$.
In other words $\det$ is the unique alternating multilinear map $\text{Mat}_n(F)\to F$ (where we interpret $\text{Mat}_n(F)$ as $n$ column vectors). The association
$$\phi\mapsto (\psi,\rho)\mapsto \det(\psi)\det(\rho)$$
is also seen to enjoy this property.