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.

Let $c_{ij}= a_i \cdot b_j$
The terms in the expansion of the determinant are 
$$ c_{i \sigma(1)} c_{2 \sigma(2)}\ldots c_{n \sigma (n)} = a_1 b_{\sigma(1)} \ldots a_n b_{\sigma(n)} = a_1\ldots a_n b_1 \ldots b_n$$
so all the terms are equal and the signs $\sum \text{sign}(\sigma)$ sum up to $0$.
