In Jeffrey Lee's differential geometry text on page 353 he defines the determinant in an interesting way using multilinear alternating maps:
Suppose $V$ is an $n$-dimensional $k$-vector space over some field $k$, and let $L^p_{\text{alt}}(V,k)$ denote the set of $p$-multilinear alternating maps from $V$ to $k$. Any $f \in \text{hom}(V,V)$ induces the pullback map $f^*: L^p_{\text{alt}}(V,k) \to L^p_{\text{alt}}(V,k)$ in the obvious way. In the case $p=n$ the dimension of $L^n_{\text{alt}}(V,k)$ is $1$, so $f^*$ acts by scalar multiplication, and we call this scalar the determinant of $f$. That is, $$f^* \omega = \det (f) \omega$$ for any $\omega \in L^n_{\text{alt}}(V,k)$.
Without using a basis, it is easy to see from this definition that $\det: \text{hom}(V,V) \to k$ is a monoid homomorphism with respect to composition in $\text{hom}(V,V)$ and multiplication in $k$. So it follows that $\det(f^{-1})=\det(f)^{-1}$ whenever $f$ is invertible.
But now my question is, how do we show without using a basis that $\det(f) \neq 0$ implies that $f^{-1}$ exists? I'm a bit rusty with my linear algebra so perhaps I'm missing something obvious.
I should probably stop trying to answer things in comments, so:
Suppose $f$ isn't invertible. Then I claim that for any $n$-form $\omega$ I have $f^*\omega = 0$. Indeed, the image of $f$ is a proper subspace of $V$, so if $v_1, \dots, v_n \in V$ then $f(v_1), \dots, f(v_n)$ are linearly dependent, so if you feed them into an alternating map you get zero.