Diagonal entries are zero, others are $1$. Find the determinant. $\det\begin{vmatrix}
0 & \cdots & 1& 1 & 1 \\
\vdots & \ddots & \vdots & \vdots & \vdots \\
1 & \cdots & 1 & 1 & 0
\end{vmatrix}=?$
Attempt:
First I tried to use linearity property of the determinants such that $$\det\binom{ v+ku }{ w
 }=\det\binom{v  }{ w
 }+k\det \binom{ u }{ w
 }$$
$v,u,w$ are vectors $k$ is scalar.
I have tried to divide it into $n$ parts and tried to compose with sense but didn't acomplish.
Second I tried to make use of "Row Reduction" i.e. adding scalar multiple of some row to another does not change the determinant, so I added the all different rows into other i.e. adding $2, 3,4,\dots,n$th row to first row and similarly doing for all rows we got 
$$\det\begin{vmatrix}
0 & \cdots & 1& 1 & 1 \\
\vdots & \ddots & \vdots & \vdots & \vdots \\
1 & \cdots & 1 & 1 & 0
\end{vmatrix}=\det\begin{vmatrix}
n-1 & \cdots & n-1& n-1 & n-1 \\
\vdots & \ddots & \vdots & \vdots & \vdots \\
n-1 & \cdots & n-1 & n-1 & n-1
\end{vmatrix}=0$$
The last determinant is zero (I guess) so the given determinant is zero?
I don't have the answer this question, so I am not sure. How to calculate this determinant?

Notice that 
$$\begin{bmatrix} 0 & 1 & \cdots & 1\\
1 & 0 & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\cdots & 0
\end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1
\end{bmatrix} = (n-1)\begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1
\end{bmatrix}$$
$$\begin{bmatrix} 0 & 1 & \cdots & 1\\
1 & 0 & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\cdots & 0
\end{bmatrix}\begin{bmatrix} 1 \\ -1 \\ 0 \\ \vdots \\ 0
\end{bmatrix} = -\begin{bmatrix} 1 \\ -1 \\ 0 \\ \vdots \\ 0
\end{bmatrix}$$
$$\begin{bmatrix} 0 & 1 & \cdots & 1\\
1 & 0 & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\cdots & 0
\end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ -1 \\ \vdots \\ 0
\end{bmatrix} = -\begin{bmatrix} 1 \\ 0 \\ -1 \\ \vdots \\ 0
\end{bmatrix}$$
$$\vdots $$
$$\begin{bmatrix} 0 & 1 & \cdots & 1\\
1 & 0 & \cdots & 1\\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 &\cdots & 0
\end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ -1
\end{bmatrix} = -\begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ -1
\end{bmatrix}$$
All eigenvectors here are linearly independent so the eigenvalues are $-1$ with multiplicity $n-1$ and $n-1$ with multiplicity one.
Therefore the determinant is $(-1)^{n-1}(n-1)$. 
