Let $B$ be a $(n-1)×(n-1)$ matrix such that: all elements on diagonal equal $0$; and all other either $1$ or $\text{-}1$.
Let $A = \begin{bmatrix}B&(1,...,1)^T\\ (1,...,1)&1\end{bmatrix}$, so $A$ be a $n×n$ derived from $B$ by adding a row and column with $1$.
What could be the rank of the matrix $A$ ?
$A$ is necessarily of full rank, since its determinant is non-zero. We can show that the determinant is non-zero by showing that it is necessarily an odd number.
Following Hans's idea here, showing that $A$ always has odd determinant is equivalent to showing that the number of permutations on $n$ objects that either have no fixed point or fix only the final entry is odd. If $d_n$ denotes the number of derangements, then we wish to show that $d_n + d_{n-1}$ is necessarily odd.
We note that $d_n$ satisfies the recurrence relation $$ d_1 = 0, \quad d_n = n d_{n-1} + (-1)^n $$ so that $d_n$ is odd iff $d_{n-1}$ is even. It follows that $d_n + d_{n-1}$ is always the sum of an even and odd number, and is therefore odd.