How to calculate the determinant of this matrix?

Question

The matrix is $\mathbf{A}=\bigl[a_{ij}\bigr]_{1\leqslant i,j\leqslant n}$ and is defined as follows:

$$a_{ij}= \begin{cases} i\; \mbox{if } i = j,\\ n\; \mbox{otherwise.} \end{cases} $$ or $$\mathbf{A}= \begin{bmatrix} 1 & n & \ldots & n\\ n & 2 & \ldots & n\\ \vdots & \vdots & \ddots & \vdots\\ n & n & \ldots & n\\ \end{bmatrix}$$

I tried one by one, for $n=1,2,3,$ and $4$ and I found a formula like:

$$\det(\mathbf{A})=(-1)^{n+1}\cdot n!.$$

I could not prove it by induction.

Answer 1

Hint: Subtract the last row from each other row, and you can get a lower triangular matrix.

Answer 2

$$L_n\leftarrow L_n-L_1$$ and you get that $$\det A=(-1)^{n-1}(n-1)\left|\begin{matrix}n&n&...&n\\n&2&...&n\\ \vdots&\vdots&\ddots&\vdots\\n&...&n-1&n\\\end{matrix}\right|=(-1)^{n-1}(-1)^{n-2}(n-1)\left|\begin{matrix}2&n&...&n\\n&3&...&n\\ \vdots&\vdots&\ddots&\vdots\\n&...&n&n\\\end{matrix}\right|=-(n-1)\left|\begin{matrix}2&n&...&n\\n&3&...&n\\ \vdots&\vdots&\ddots&\vdots\\n&...&n&n\\\end{matrix}\right|=(-1)^2(n-1)(n-2)\left|\begin{matrix}3&n&...&n\\n&4&...&n\\ \vdots&\vdots&\ddots&\vdots\\n&...&n&n\\\end{matrix}\right|=...=(-1)^n n!.$$