Let $P$ be a $n$-gon with vertices $(x_1,y_1),\ldots,(x_n,y_n)$ enumerated clockwise. Then the area $\text{Area}(P)$ of $P$ is
$$ \text{Area}(P) = \sum_{i=1}^n\frac{1}{2}(x_{i+1}-x_i)(y_{i+1}+y_i).$$
What is the easiest way to prove this?
I think one way would be to use induction on the number of vertices. For the inductive step, one could use the fact that every $n$-gon for $n>3$ has a diagonal.
On
You may just compute the (oriented) area of a triangle with vertices at $(0,0),(x_i,y_i),(x_{i+1},y_{i+1})$ and add the contributes to get: $$ A = -\frac{1}{2}\sum_{i=1}^{n}\left(x_i y_{i+1} - x_{i+1} y_i\right)=\frac{1}{2}\sum_{i=1}^{n}(x_{i+1}-x_i)(y_{i+1}+y_i) $$ where the last equality follows from $\sum_{i=1}^{n}\left(x_{i+1}y_{i+1}-x_i y_i\right)=0$.
For short, your formula is equivalent to the shoelace formula.
Use Green's area formula $${\rm area\,}(P)={1\over2}\int_{\partial P}(x\>dy-y\>dx)$$ and parametrize the segments $\sigma_k$ making up $\partial P$ as $$\sigma_k:\quad t\mapsto \bigl((1-t)x_k+t x_{k+1}, \>(1-t)y_k+t y_{k+1}\bigr)\qquad(0\leq t\leq1)\ .$$