Understanding determinant function Suppose $\dim_K V=d$. Then, there is a unique such $f$ satisfy $f(e_1,...,e_n)=1$. We call this $f$ the determinant function.
I think, it is a strong theorem but I couldn't interiorise what the theorem say. Can you explain clearly? Why $f(e_1,...,e_n)=1$? And why unique?
 A: Let $V$ and $M$ be vector spaces over the field $K$. It is well-known that any multilinear, alternating function $f: V^{\times m} \to M$ must factor through what is called the $m$th exterior power of $V$, denoted $\bigwedge^m V$, so $f$ may be written as a composition $V^{\times m} \to \bigwedge^m V \to M$, where the last map is linear. Moreover, this factorisation is unique.
The dimension of $\bigwedge^m V$ is $\binom{\dim V}{m}$, and so when $m = \dim V$ this space is 1-dimensional. An implication of this is that a multilinear alternating map in $\dim  V$ variables is the same as a linear map between the 1-dimensional spaces $\bigwedge^{\dim V} V \to K$, and so it's essentially just a scalar.
There are different scalars though: if you put a factor of 2 in front of the determinant formula, it would still satsify multilinear, alternating, and nonzero. So requiring that $f(e_1, \ldots, e_n) = 1$ is just a way of scaling it to something standard.
You can find out more about what I've mentioned here by looking for notes on "multilinear algebra" and "exterior powers", or "exterior algebra". It's probably not the most basic way of seeing these facts, but I think it gives the best understanding.
