I know the definition of Determinant function that it is a mapping $D: \mathbb{K}^{n \times n} \rightarrow \mathbb{K}$ such that (i) $D$ is n-linear (ii) $D(A) = 0$, if two rows are equal (iii) $D(I) = 1$ for the identity matrix I. where $\mathbb{K}$ is a commutative ring and $\mathbb{K}^{n \times n}$ is the set of all $n \times n$ matrices.
Can anyone tell me by using the above definition of determinant function how to prove $det(AB) = det(A) det(B)$ and $det(A) = det(A^T)$?
Here is a sketch: fix $A$ and define $D(B) = detAB$. It is easy to show that $D$ preserves addition and scalar multiplication in the columns of $B$, and that D is alternating in the columns of $B$. Therefore, by uniqueness of the alternating, multilinear function, we must have $D(B)=D(I)\ detB=\ detAI\ detB=\ detA\ detB$.
Now use the $QR$ factorization of $A^T$ and what we have just proved, to show the second claim.
Yet another way would be to use your definition to show that the $det$ function $must$ have the form
$\displaystyle \sum_\lambda \operatorname{sgn} \left({\lambda}\right) b_{1 \lambda \left({1}\right)} b_{2 \lambda \left({2}\right)} \cdots b_{n \lambda \left({n}\right)}$ where the sum is taken over all permutations $\lambda:\left \{ 1,\cdots, n \right \}\to \left \{ 1,\cdots, n \right \}$
and from here use the arguments you are probably used to.