Shortest and most elementary proof that the product of an $n$-column and an $n$-row has determinant $0$ Let $\bf u$ be any column vector and $\bf v$ be any row vector, each with $n \geq 2$ arbitrary entries from a field. Then it is well known that 

${\bf u} {\bf v}$ is an $n \times n$ matrix such that $\det({\bf uv})=0$. 

I am curious to know the $\bf shortest$ and $\bf most~elementary$ proof of this result, say understandable by a (good) high school student. I have one in mind, but to make this  interesting, perhaps I should let you present your version first?
UPDATE: Someone already presented (below) the same proof I had in mind. But let's see if there is a simpler proof; finding one is the main motivation here.

This question can be easily answered by recalling determinant properties. Specifically:
$det(A_1,A_2,\dots,cA_j,\dots,A_n)$ = $c$$(det(A_1,A_2,\dots,A_j,\dots,A_n))$
Therefore if $u = \begin{bmatrix}u_1\\ u_2\\ \vdots \\ u_n \end{bmatrix}$ and $v = \begin{bmatrix} v_1,& v_2,& \dots,& v_n\end{bmatrix}$
Then:
$uv$ = $\begin{pmatrix} u_1v_1 & u_1v_2 & \cdots & u_1v_2 \\
  u_2v_1 & u_2v_2 & \cdots & u_2v_n \\
  \vdots  & \vdots  & \ddots & \vdots  \\
  u_nv_1 & u_nv_2 & \cdots & u_nv_n
 \end{pmatrix} $
Then, by the property mentioned, $det(uv)$ = $v_1v_2$$\cdots v_n$ $det$ $\begin{pmatrix} u_1 & u_1 & \cdots & u_1 \\
  u_2 & u_2 & \cdots & u_2 \\
  \vdots  & \vdots  & \ddots & \vdots  \\
  u_n & u_n & \cdots & u_n
 \end{pmatrix} $
Clearly this determinant is zero, therefore $det(uv)$ = $0$.
