1. **Problem Statement:** Prove that for matrices $A$ and $B$, the transpose of their product satisfies $$(AB)^T = B^T A^T.$$ 2. **Definitions and Setup:** Let $A$ be an $m \times n$ matrix with elements $a_{ij}$ and $B$ be an $n \times p$ matrix with elements $b_{ij}$. The product $C = AB$ is an $m \times p$ matrix with elements $$c_{ij} = \sum_{k=1}^n a_{ik} b_{kj}.$$ The transpose of a matrix $A$, denoted $A^T$, is defined by swapping rows and columns: $$a_{ij}^T = a_{ji}.$$ 3. **Expressing the transpose of the product:** The transpose of $C = AB$ is a $p \times m$ matrix with elements $$(AB)^T_{ij} = c_{ji}.$$ Using the definition of $c_{ji}$, we have $$c_{ji} = \sum_{k=1}^n a_{jk} b_{ki}.$$ 4. **Expressing $B^T A^T$:** The transpose matrices are $B^T$ of size $p \times n$ with elements $b_{ij}^T = b_{ji}$ and $A^T$ of size $n \times m$ with elements $a_{ij}^T = a_{ji}$. Their product $B^T A^T$ is a $p \times m$ matrix with elements $$ (B^T A^T)_{ij} = \sum_{k=1}^n b^T_{ik} a^T_{kj} = \sum_{k=1}^n b_{ki} a_{jk}.$$ 5. **Comparing the two expressions:** From steps 3 and 4, we see that $$ (AB)^T_{ij} = \sum_{k=1}^n a_{jk} b_{ki} = \sum_{k=1}^n b_{ki} a_{jk} = (B^T A^T)_{ij}.$$ 6. **Conclusion:** Since the elements of $(AB)^T$ and $B^T A^T$ are equal for all $i,j$, the matrices are equal: $$ (AB)^T = B^T A^T.$$ This completes the proof.