Let $A$ be a square matrix of size $n$ with $$a_{ij} = \begin{cases} \phantom{+}1 & \text{if } i<j-1, \\ \phantom{+}0 & \text{if } i=j-1, \\ -1 & \text{if } i>j-1.\end{cases}$$ I'm asked to find the determinant of $A$ with the hint of using induction. I check some values: $$n=2 \implies |A| = 1$$ $$n=3 \implies |A| = -1$$ $$n=4 \implies |A| = 1$$ I seems that $|A| = (-1)^n$ (as a side note I know this is true).
So let's assume $n=2$ is my base case and let $|A|_n$ denote determinant of matrix $A$ of size $n$.
I need to prove that $|A|_{n+1} = (-1)\cdot|A|_n$, but this is where I get stuck. I've tried using expansion along a row, $$|A| = \sum_{j}a_{ij}A_{ij}$$ which seems promising if I start at the bottom right ($a_{nn} = -1$ for all $n$ and $A_{ij}$ is the previous determinant). But I can't find a pattern for the rest of that sum, the rest should equal zero.
How can I prove this using induction and beginner matrix theory?
$$ A_n=\left(\begin{array}{ccccc|c} -1& 0& 1& \cdots&\cdots&1\\ -1&-1& 0& \ddots&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots&0&1\\ -1&\cdots&\cdots&\cdots&-1&0\\ \hline -1&\cdots&\cdots&\cdots&-1&-1 \end{array}\right) =\pmatrix{A_{n-1}&u\\ -\mathbf1^T&-1}. $$ Subtract the last row of $A_n$ by the last but one row, we get $\det(A_n)=\det\pmatrix{A_{n-1}&u\\ 0&-1}=-\det(A_{n-1})$.