Determinant of a non-square block matrix $M_{n\times k}$ is defined as a matrix whose all elements are '-1'.
The following block matrix is as such:
$A =\begin{bmatrix}m\cdot I_{n-1} & M_{n-1\times m}\\M_{m\times n-1} & n\cdot I_{m}\end{bmatrix}$
prove the following: $det A = n^{m-1}\cdot m^{n-1}$
 A: Let more generally let $C(\gamma) = \gamma 1_{n-1}1_{m}^T $ denote the ${n-1}\times m$ matrix with each element equal to $\gamma$ (here $1_k$ denotes the k-dimensional column vector of all ones) and let $A(\gamma) = \begin{pmatrix} mI_{n-1} & C(\gamma) \\ C(\gamma)^T & nI_m\end{pmatrix}$. Your problem is to  compute the determinant of $A(-1)$.
Using the following identity
$$ \begin{pmatrix} I_{n-1} & 0 \\ - \dfrac{C(\gamma)^T}{m} & I_m \end{pmatrix} \begin{pmatrix} mI_{n-1} & C(\gamma) \\ C(\gamma)^T & nI_m\end{pmatrix} = \begin{pmatrix} mI_{n-1} & C(\gamma)\\ 0 & nI_m - \dfrac{C(\gamma)^TC(\gamma)}{m} \end{pmatrix}.$$ 
we have on taking determinants $$
\begin{align}
\operatorname{det}(A(\gamma)) &=\operatorname{det}(mI_{n-1})\operatorname{det}(nI_m - \dfrac{C(\gamma)^TC(\gamma)}{m})\\ &= m^{n-1} \times n^{m} \det(I_m  - \dfrac{C(\gamma)^TC(\gamma)}{mn})\\ &= m^{n-1}n^{m} \det(I_{m} - \gamma^2 \dfrac{(n-1)}{nm} 1_{m}1_{m}^{T})  \tag{*}
 \\
&= m^{n-1} n^m \det(I_m - \delta 1_m 1_m^T)
\end{align}
$$
where $\delta = \gamma^2\dfrac{n-1}{nm}.$
In (*) we have used $C(\gamma) = \gamma 1_{n-1}1_{m}^T$.
Consider the following identities:
$$
\begin{pmatrix} I_m & 0 \\ -1_m^T & 1 \end{pmatrix} \begin{pmatrix} I_m & \delta 1_m \\ 1_m^T & 1 \end{pmatrix} = \begin{pmatrix} I_m & \delta 1_m \\ 0 & 1 - m \delta \end{pmatrix},$$
$$\begin{pmatrix} I_m & -\delta 1_m \\ 0 & 1 \end{pmatrix}  \begin{pmatrix} I_m & \delta 1_m \\ 1_m^T & 1 \end{pmatrix} = \begin{pmatrix} I_m - \delta 1_m 1_m^T & 0 \\ 1_m^T & 1 \end{pmatrix}. $$
On taking determinants we have $\det\begin{pmatrix} I_m - \delta 1_m 1_m^T \end{pmatrix} = 1 - m \delta.$
This gives $\det(A(\gamma))= m^{n-1}n^m(1 - m \delta) = m^{n-1}n^m (1 - \gamma^2\dfrac{n-1}{n})$ so $ \det(A(-1))=m^{n-1}n^{m-1}.$
