Showing the Weyl element is sent to its negative by a certain element of the Weyl group

Question

Let $\mathfrak g$ be a complex semi-simple Lie algebra with a choice of positive root system. Let $\rho$ be the Weyl element, i.e. the sum of the fundamental weights (or half the sum of the positive roots). There is a unique element $w_0$ of the Weyl group that sends the positive Weyl chamber to the negative one. From the classification of such algebras, I can see that $w_0 \rho = -\rho$. However, what is a simple (i.e. short) proof of this?

Answer 1

The number of positive roots sent to negative roots by $w \in W$ is equal to the length, $\ell(w)$, of $w$. In the case of $w_0$, the length $\ell(w_0)$ is the maximal value, equal to the total number of positive roots.

Use the definition of $\rho$ as half the sum of all positive roots and this fact to conclude that $w_0 \rho = -\rho$.