Trivial zeroes of Zeta are simple The trivial zeroes of the Riemann Zeta function are located on $-2\mathbb N^*$ and they are simple. It is not difficult to see that, but the proof I have in mind is using the fact that $\xi(-2k)=\xi(1+2k)>0$ for $k\ge 1$ integer.
Well, it is not so simple, since it is using (part of) the functional equation for $\xi$, definitely  not an elementary fact. Is there a simpler proof of the assertion in the title?

A priori, the Riemann zeta function only makes sense for $\text{Re} (s) > 1$. Most commonly, one understands the zeta function for $\text{Re} (s) < 0$ through some sort of functional equation. One does not need the reflecting functional equation $\xi(s) = \xi(1-s)$. Riemann's original functional equation
$$ \zeta(s) = 2^s \pi^{s-1} \sin(\pi s/2) \Gamma(1-s) \zeta(1-s)$$
is sufficient. Another common functional equation is
$$ \zeta(s) \Gamma(s) = \int_0^\infty  \frac{x^s}{e^x - 1} \frac{dx}{x},$$
which highlights the relationship between the Bernouilli numbers and the zeta function. This functional equation (with some consideration of the zeroes and poles involved) is mentioned in Daniel Fischer's answer here.
Of course, there are very many ways to meromorphically continue the zeta function. One would expect most of the good ways to all illustrate the locations of the zeroes.
