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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?

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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 $\zeta'(-2m) \neq 0$.

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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.

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    $\begingroup$ this is not entirely true: there is expression for the zeta function that converge for all value of $z$ and which can substitute to the functional equation to prove the analytic continuation. $\endgroup$
    – Simon Henry
    Commented Aug 28, 2015 at 15:39
  • $\begingroup$ @Igor Rivin It is indeed possible to define the Riemann Zeta function on the whole complex plane as a meromorphic function with a single simple pole at 1, using a variation of Euler-MacLaurin formula, much simpler to handle than the functional equation. $\endgroup$
    – Bazin
    Commented Aug 28, 2015 at 15:51
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    $\begingroup$ @Bazin: You should include your working definition of the zeta function in the question, as any answer will crucially depend on it. $\endgroup$
    – Emil Jeřábek
    Commented Aug 28, 2015 at 16:18
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    $\begingroup$ In section 6, chapter 16 of Ireland & Rosen's book they show how to extend the zeta-function to the half-plane $\{s : {\rm Re}(s) > -k\}$ for $k = 0, 1, 2, \dots$ step by step. No functional equation is used or even arises in this method. (Whether this method can be carried out more carefully to tell you that a trivial zero is a simple zero is not something I checked.) When Gerry writes that you must use the functional equation to define the zeta-function on all of $\mathbf C$ he is incorrect. $\endgroup$
    – KConrad
    Commented Aug 28, 2015 at 19:00
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    $\begingroup$ @KConrad, the method in Ireland and Rosen's book seems to be more or less the same as the second method in my answer. $\endgroup$
    – Peter Humphries
    Commented Aug 28, 2015 at 19:47

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