Trivial zeroes of the Riemann Zeta function 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: As implicitly stated in Igor Rivin's answer, the standard way to show the vanishing of $\zeta(s)$ at negative even integers, and that such zeroes are simple, would be to use the functional equation, as this is the standard method to define $\zeta(s)$ to the left of the line $\Re(s) = 0$.
However, there are other methods that give a meromorphic continuation of $\zeta(s)$ to $\mathbb{C}$, and from which one can show that $\zeta(s)$ has simple zeroes at the negative even integers. Here are papers describing two such methods:
http://www.ams.org/journals/proc/1994-120-02/S0002-9939-1994-1172954-7/
https://www.math.ucdavis.edu/~tracy/courses/math205A/EulerMaclaurinSummation.pdf
The second method is quite famous. It is quite standard that one can use partial summation to show that
\[\zeta(s) - \frac{1}{s - 1} = 1 - s \int_{1}^{\infty} \frac{\{x\}}{x^s} \, \frac{dx}{x}\]
for $\Re(s) > 1$, with the right-hand actually defining a holomorphic function for $\Re(s) > 0$. This gives a meromorphic continuation of $\zeta(s)$ to the open half-plane $\Re(s) > 0$ with a simple pole at $s = 1$.
A generalisation of partial summation is Euler-Maclaurin summation, and the second link above shows how using this, one can extend $\zeta(s)$ meromorphically to the open half-plane $\Re(s) > -k$ for any nonnegative integer $k$. Moreover, using well-known properties of Bernoulli numbers, one can read off from these that $\zeta(s)$ vanishes when $s = -2m$ for some positive integer $m$, while one can differentiate term-by-term to show that $\zeta'(-2m) \neq 0$. 
A: As pointed out by Gerry Myerson in the answer to this question, you cannot even define zeta for negative real parts without using the functional equation, so no.
