linear algebra - Prove $det(A - nI_n) = 0$.

Problem: Prove that $\det(A - n I_n) = 0$ when $A$ is the $(n \times n)$-matrix with all components equal to $1$.

Attempt at solution: I tried to use Laplace expansion but that didn't work. I see the matrix will be of the form $$\begin{align*} \begin{pmatrix} 1-n & 1 & \cdots & 1 \\ 1 & 1-n & \cdots & 1 \\ \vdots \\ 1 & 1 & \cdots & 1-n \end{pmatrix} \end{align*}$$ I want to somehow get two equal rows or columns here, or a row/column of zero using elementary operations. But I don't see what I should do?

Answer

Hint: Add all other rows to the first row. What row you'll obtain after that?