TITLE: Determinant of a specific $n\times n$ matrix
QUESTION [5 upvotes]: Let $M_n=
  \begin{bmatrix}
    1 & 2 & 3 & ... & n\\
    2 & 2 & 3 & ... & n\\
    3 & 3 & 3 & ... & n\\
    \vdots & \vdots & \vdots & \ddots & n\\
    n & n & n & n & n
  \end{bmatrix}$
What's $\det(M_n)$ ?
Seems like it is one of these exercises virtually impossible to do without the one trick, which is why I ask this question.
Computing the first few determinants yields:
$\det(M_1)=1$
$\det(M_2)=-2$
$\det(M_3)=3$
so if I were to make a wild guess, I would say $\det(M_n)=n(-1)^{n+1}$
The induction process didn't work too well for me because I can't find the determinant once a row and a column have been added to pad the matrix to dimension $n+1$.
I tried to use the well-known formula for the determinant by cutting the inside in four parts ($M_n$ in the upper-left, $(n+1)$ in the lower-right, and the $(n+1)$ row and column, but at the end I must compute the determinant of a sum of matrices which didn't help...
I don't know the name of this specific matrix, so maybe this is a repost, but if this is the case, I can't find the original one
I'd be grateful if someone could provide some hints.

REPLY [3 votes]: Multiply $\frac{n}{n-1}$ to the $(n-1)$-th row and subtract it to the $n$th row. Then the matrix will be in form of
$\pmatrix{M_{n-1}&n\mathbf{1}\\
\mathbf{0}&-\frac{n}{n-1}}$
where $\mathbf{1}=(1, 1, \cdots, 1)_{n-1}^T, \mathbf{0}=(0, 0, \cdots, 0)_{n-1}$. Thus inductively we have $\det(M_n)=-\frac{n}{n-1}\det(M_{n-1})$.