The matrix has 0's along the main diagonal, 1's everywhere else? Then $\displaystyle (-1)^n(n- 1)$ is NOT correct. When n= 2, the matrix is $\displaystyle \begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}$ which has determinant -1, not $\displaystyle (-1)^2(2- 1)= 1$. Perhaps it is $\displaystyle (-1)^{n-1}(n-1)$.
One thing I notice about it is that given $n$ the eigenvalues are $n-1$, and $n-1$ repeated values of $-1$
Given this clearly the determinant is just $(-1)^{n-1}(n-1)$
it's not too hard to see that the eigenvector of all 1's produces the eigenvalue of $n-1$, it's just the sum of the row values of $C$ which is clearly $n-1$
the eigenvectors corresponding to the eigenvalue of $-1$ are of the form
$(-1, 0, 0, 0, \dots , 1, 0, \dots, 0)$
where the $1$ appears in $n-1$ different rows corresponding to $n-1$ different eigenvectors.
I know this doesn't solve your problem but it might help.
cheers