Without choosing bases, how to show that the determinant is multiplicative in this sense? 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.?)
 A: The lateness of this answer gives a new meaning to the word "overdue". In any case, I noticed this question while browsing this website and I thought I'd answer it because that's what you do when you're faced with a question right?
I'm not sure if this is what you're looking for; it's basically equivalent to your basis-dependent proof above and it's based on the section of the Wikipedia article: http://en.wikipedia.org/wiki/Exterior_algebra#Direct_sums .
The short exact sequence of vector spaces:
(1) $0\to W\to V\to V/W\to 0$
results in a natural isomorphism:
(2) $\bigwedge^{k} V\cong \bigwedge^{i} W\otimes \bigwedge^{j} V/W$
where $k$, $i$, and $j$ are the dimensions of $V$, $W$, and $V/W$, respectively. The endomorphism of (1) (in the category of short exact sequences of vector spaces) given by the triple $(\psi,\phi,\rho)$ induces an endomorphism of (2) (i.e., a commutative diagram!). The commutativity of this commutative diagram (I hope this isn't bad grammar!) combined with the characterisation of the determinant via the top exterior power results in the equation $\det \phi = \det \psi \det \rho$.
Does that answer your question? The only thing I've done really is inserted fancy terms like "isomorphism", "short exact sequence", and "exterior power" in your proof ...
Edit: The isomorphism (2) can be derived by using the fact that (1) splits; see the comments below. Of course, (1) splits because every object in the abelian category of vector spaces over a field is free. In some sense, this is using the existence of a basis for each finite dimensional vector space (choosing a splitting of (1) is equivalent to choosing a basis of the vector space $V/W$). So, you might object that this proof is "nasty" because it involves choices (and I agree). 
Ultimately, I suppose the question is to "hide" any use of choices of bases because this is what other apparently "basis-independent" proofs in linear algebra do. The approach is to introduce a construction which "collects" all choices. (If there's no natural choice in mathematics, then you just take them all! Sadly, this is something we can't do in real life.) In some sense, this is what the exterior algebra characterization of the determinant does.
A: 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.
A: In geometric (or Clifford) algebra, the determinant of a linear map from vectors to vectors is obtained through "outermorphism".  Given a linear map $\underline \phi$, there is a natural extension of this map to arbitrary $k$-vectors given by
$$\underline \phi(a_1 \wedge a_2 \wedge \ldots \wedge a_k) = \underline \phi(a_1) \wedge \underline \phi(a_2) \wedge \ldots \wedge \underline \phi(a_k)$$
for $k$ distinct vectors $a_1, a_2, \ldots, a_k$.
An $n$-dimensional vector space has a pseudoscalar that characterizes it.  This can be formed by taking wedge products of basis vectors.  However, as the vector space of pseudoscalars is 1-dimensional, all pseudoscalars are scalar multiples of all others.  The pseduoscalar is an $n$-vector.
Choose some pseudoscalar for the space and denote it $i$.  Then the determinant is the number $f$ such that
$$\underline \phi(i) = f i$$
Again, it follows that there is only one such number because $i$ is a member of a 1d vector space.
In GA parlance, you're talking about a linear map being decomposed into eigenblades.  There is a blade $W$ and its complement $W^\perp = iW^{-1}$.  You know that $\underline \phi(W) = \alpha W$ for some scalar $\alpha$.
Define a projection map $\underline P(a)$ such that $\underline P(a) \wedge W = 0$.  Similarly, there is a rejection map $\underline P^\perp(a)$ such that $\underline P^\perp(a) \wedge W^\perp = 0$.  Of course, $\underline P + \underline P^\perp = \underline I$, the identity.
Break down the original map $\underline \phi$ as follows:
$$\underline \phi(a) = [\underline P + \underline P^\perp] \underline \phi[ \underline P + \underline P^\perp](a) = [\underline{P \phi P} + \underline{P^\perp \phi P^\perp} + \underline{P \phi P^\perp} + \underline{P^\perp \phi P}](a)$$
The first two terms are the maps $\underline \psi$ and $\underline \rho$ that you defined, so the breakdown is
$$\underline \phi(a) = \underline \psi(a) + \underline \rho(a) + \underline{P^\perp \phi P}(a) + \underline{P \phi P^\perp}(a)$$
Now, we said quite specifically that $\phi(W) = \alpha W$.  That means that third term, $\underline{P^\perp \phi P}$, is exactly zero (the zero block you observed in matrix form).  Now we can take the full determinant by seeing that $i = W^\perp \wedge W$.
$$\underline \phi(i) = \underline \phi(W^\perp) \wedge \underline \phi(W) = [\underline \rho(W^\perp) + \underline{P \phi}(W^\perp)] \wedge [\underline \psi(W)] = [\underline \rho(W^\perp) + \underline{P \phi}(W^\perp)] \wedge [\alpha W]$$
Now, we'll get a term that is $\underline{P \phi}(W^\perp) \wedge [\alpha W]$.  This is zero, as the projection $\underline P$ will force that term to lie within the subspace of $W$, and $W \wedge W= 0$ always.  This leaves us with
$$\underline \phi(i) = [\underline \rho(W^\perp)] \wedge [\underline \psi(W)]$$
Identity $W^\perp$ acts like the pseudoscalar for $\underline \rho$, just as $W$ acts as a pseudoscalar for $\underline \psi$. The result is
$$\underline \phi(i) = (\det \underline \rho) W^\perp \wedge (\det \psi) W = (\det \underline \rho)(\det \underline \psi) W^\perp W$$
And again, $W^\perp W = i$, so we're done.  I think the big key is using the projection/rejection maps to get a basis-independent statement of the block matrix form.  The Clifford algebra means of talking about determinants makes it more obvious that you can take the determinants of these restricted maps because you need only supply the correct pseudoscalar for the map to get its determinant.  It also lets us break down the determinant as the wedge of $W^\perp$ and $W$ in a basis independent way.
To carry out this calculation, you do have to choose some pseudoscalar $i$, but this is much relaxed compared to choosing a basis.  You could always choose a pseudoscalar that is unit, for instance.
