The Question:
Let $M$ be an $n \times n$ matrix and suppose that $M =\begin{bmatrix}I&B\\0&A\end{bmatrix} ,$ where $A$ is a $j \times j$ matrix with $j < n$; where $I$ is the $(n − j) \times (n − j)$ identity matrix; and where $0$ a matrix of zeros. Prove that $\det(M) = \det(A)$.
I would like to prove this using Leibniz's formula of the determinant: For an $n \times n$ matrix $A$,
$$\det(A)=\sum_{\sigma\in S_n} \operatorname{sgn}(\sigma)\cdot a_{1,\sigma(1)}\cdot...\cdot a_{n,\sigma(n)} = \sum_{\sigma\in S_n} \operatorname{sgn}(\sigma)\cdot \prod_{i=1}^na_{i,\sigma(i)}$$
Any help would be appreciated, thank you.