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.

Each row of $uv$ is a multiple of $v$. If the first row is all zeros, the determinant is zero. If not, then the second row is a multiple of the first. The determinant is unchanged if we subtract that multiple of the first row from the second row. The second row is now all zeros and so the determinant is zero.
