Using permutations to find the determinant. I came across the following problem

Let matrix A be a $n\times n$ square matrix such that $a_{ij}$ = max{ i , j }. Prove that det(A) = $(-1)^{n-1}n$

I have read the post If $a_{ij}=\max(i,j)$, calculate the determinant of $A$ and understood all the answers adressing the same problem.
I just wanted to see an alternative method which does not use row subtractions.
As, the determinant fromulla $n(-1)^{n-1}$ is reminiscent of $\frac{d}{dx}(x^{n})$ Is there some way we could use differentiation of determinant technique? How would it be if one goes by the formulla of determinant in terms of permutations?
 A: One alternative approach:
$$
A_{n+1}=
\begin{pmatrix}
1 & 2 & 3 & \cdots &n & n+1 \\
2 & 2 & 3 & \cdots & n & n+1 \\
3&3&3&\cdots &n & n+1\\
\vdots & \vdots &\vdots &\ddots&\vdots & \vdots\\ 
n & n & n&\cdots& n & n+1 \\
n+1 & n+1 & n+1&\cdots& n+1 & n+1
\end{pmatrix}.
$$
Subtract $\frac{n+1}{n}$ times the second to last row from the last row to get
$$
\begin{pmatrix}
1 & 2 & 3 & \cdots &n & n+1 \\
2 & 2 & 3 & \cdots & n & n+1 \\
3&3&3&\cdots &n & n+1\\
\vdots & \vdots &\vdots &\ddots&\vdots & \vdots\\ 
n & n & n&\cdots& n & n+1 \\
0 & 0 & 0 &\cdots & 0 & -\frac{n+1}{n}
\end{pmatrix}.
$$
Now, because of the block triangular structure, we have
$$
\det(A_{n+1}) = - \frac{n+1}{n}\det(A_n).
$$
A: I think if you subtract column $j$ to column $j+1$ for all columns you get an upper triangular matrix with all $1$ on diagonal and can develop determinant easily since on last line all but one term are zeroes.
$$
        \begin{pmatrix}
        1 & 1 & 1 & \cdots &1 \\
        2 & 0 & 1 & \cdots & 1 \\
        3&0&0&\cdots &1\\
        \vdots & \vdots &\vdots &\ddots&\vdots\\ 
        n & 0 & 0&\cdots& 0 \\
        \end{pmatrix}
=\begin{pmatrix}[i]&T_{n-1}\\
        n&0
        \end{pmatrix}
$$
with $\det(T_{k})=1$.
