Here is a natural path to the idea of determinant (though this is not how they were originally developed).

An *alternating $k$-linear function* on a vector space $V$ over a field $\Bbb F$ is a map $f\,:\, V^k \to \Bbb F$ which is

- Linear in each argument: $$f(v_1, \ldots, v_{i-1}, av_i + bw_i, v_{i+1}, \ldots, v_k) = af(v_1, \ldots, v_i, \ldots, v_k) + bf(v_1, \ldots, w_i, \ldots, v_k)$$ for all $i$.
- Changes sign under exchange of any two arguments: $$f(v_1, \ldots, v_i, \ldots, v_j, \ldots v_k) = -f(v_1, \ldots, v_j, \ldots, v_i, \ldots v_k)$$ for all $i \ne j$

It is easy to see that if $f,g$ are two alternating $k$-linear functions on $V$, then so is $af + bg$ for any $a,b \in \Bbb F$, so the alternating $k$-linear functions on $V$ form another vector space $A^k(V)$. Some development shows that if $V$ has dimension $n$, then $A^k(V)$ has dimension $n \choose k$. In particular $A^n(V)$ has dimension $1$.

Now if $M\,:\,V \to V$ is linear and if $f\in A^k(V)$, then the map $$M_kf\,:\, V^k \to \Bbb F\,:\,(v_1, ... v_k) \mapsto f(Mv_1, ..., Mv_k)$$ is also alternating $k$-linear. And clearly $M_k(af+bg) = aM_kf + bM_kg$, so $M_k$ defines a linear map from $A^k(V)$ to itself (i.e., an endomorphism of $A^k(V)$).

Since $A^n(V)$ is one dimensional, any endomorphism is just multiplication by some element of the field $\Bbb F$. Thus we define the determinant of $M$ to be the unique element $\det(M) \in \Bbb F$ such that $$M_nf = \det(M)f\text{ for all }f \in A^n(V)$$

All the properties of determinants, including the permutation formula can be developed from this. Certain properties of determinants that are difficult to prove from the Liebnitz formula are almost trivial from this definition. In particular that $\det(MN) = \det(M)\det(N)$.

There is a close connection between the space of alternating $k$-linear functions and the $k$-order wedge product of a space, so I could have very similarly developed the determinant based on the wedge product, but alternating $k$-linear functions are easier conceptually.