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?

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
$$
\det(A_{n+1}) = - \frac{n+1}{n}\det(A_n).
$$
