Show that the determinant of the matrix $E$ is $0$

Let $$E=\begin{pmatrix} (1-n)e_1 & e_2 & ... & e_n\\ e_1 & (1-n)e_2 & ... &e_n\\ . & . & . &.\\ . & . & . &.\\ e_1 & e_2 & ... & (1-n)e_n \end{pmatrix}$$ an $$n \times n$$ matrix. I need to prove that the determinant is zero. I thought about using guassian elimination, and somehow $$e_{nn}$$ (position) will lead to zero. So, as a triangular matrix with a zero in the diagonal means $$Det(E)=0$$, then it is solved.

However, I would like to know if there's a better way to prove it, or a hint, and if what I thought is correct.

First $$(1)$$, factor out the coefficients. Second $$(2)$$, add all rows to the first row: $$\begin{vmatrix} (1-n)e_1 & e_2 & ... & e_n\\ e_1 & (1-n)e_2 & ... &e_n\\ . & . & . &.\\ . & . & . &.\\ e_1 & e_2 & ... & (1-n)e_n \end{vmatrix}$$$$\overset 1=(e_1\cdots e_n)\begin{vmatrix} (1-n) & 1 & ... & 1\\ 1 & (1-n) & ... &1\\ . & . & . &.\\ . & . & . &.\\ 1 & 1 & ... & (1-n) \end{vmatrix}$$$$\overset 2 =(e_1\cdots e_n)\begin{vmatrix} (1-n)+(n-1) & (1-n)+(n-1) & ... & (1-n)+(n-1)\\ 1 & (1-n) & ... &1\\ . & . & . &.\\ . & . & . &.\\ 1 & 1 & ... & (1-n) \end{vmatrix} \\= (e_1\cdots e_n)\begin{vmatrix} 0 & 0 & ... & 0\\ 1 & (1-n) & ... &1\\ . & . & . &.\\ . & . & . &.\\ 1 & 1 & ... & (1-n) \end{vmatrix}$$$$=0$$
• Nice way to write $(e_1,..,e_n)$, it makes it easier, thanks. Nov 6 '20 at 17:38
Consider the vector $$v = [1,1,\ldots, 1]^T$$. You can calculate component wise that $$E^Tv = 0$$ which implies that $$E$$ is singular so $$\det(E^T) = 0$$ and since $$\det(E)=\det(E^T)$$ we have that $$\det(E)=\det(E^T) = 0$$ as desired.