Multiplicativity of determinants

$$|AB|=|A||B|$$ Proof 1: direct computation (The most elementary proof)
Let $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ and $B=\begin{pmatrix}x&y\\z&t\end{pmatrix}$. Then $AB=\begin{pmatrix}ax+bz&ay+bt\\cx+dz&cy+dt\end{pmatrix}$. We thus have $$\begin{align}|AB|&=(ax+bz)(cy+dt)-(ay+bt)(cx+dz)\\ &=acxy+adxt+bcyz+bdtz-acxy-bcxt-adyz-bdzt\\ &=(ad-bc)xt-(ad-bc)yz\\ &=(ad-bc)(xt-yz)\\ &=|A|\cdot |B| \quad \Box \end{align}$$ Proof 2: exterior algebra / wedge product
$|AB|=e_1\wedge \cdots \wedge e^n=|A|(Be_1\wedge \cdots \wedge e^n)=|A|\cdot |B|(e_1\wedge \cdots \wedge e_n) \;\Box$

Proof 3: elementary row operations
Suppose A is singular. Then $|A|=0 \Rightarrow |A||B|=0$. On the other hand, $|AB|=0$ since $\text{rank}(AB)\leq\text{rank}(A)$. Thus $|AB|=|A||B|=0$.

Suppose A is invertible. Then $A$ is a product of elementary matrices $A=E_1E_2\cdots E_k$. Thus $AB=E_1E_2\cdots E_kB$. Then $|AB|=|E_1||E_2|\cdots|E_k||B|=|A||B|. \quad \Box$

More to explore
More proofs
Proof by analysis -- fundamental algebra of algebra