Trivial zeros of Zeta Riemann Function

[Facktor]

According to Wikipedia, the Zeta Riemann Function is defined as follows:

\begin{equation}
\zeta(z) = \sum_{k=1}^{\infty}\frac{1}{k^{z}}, \forall z \in \mathbb{C}, Re[Z] > 1.
\end{equation}
Well, the trivial zeros are the negative even numbers. Is that a consequence of the following expression?
\begin{equation}
\zeta(-n) = -\frac{B_{n+1}}{n+1}, n \geq 1, n \in \mathbb{I},
\end{equation}
where B is a Bernoulli number.

But the above expression is a definition of the Zeta Function for negative integers? Why?

Thanks!

 

[RamaWolf]

The values of the Zeta-function for negative integers is a consequence of the 'fnctional equation' of the Zeta-function. From this 'functional equation' it turns out that:

(1) Zeta(-n) = 0 if n is even
(2) Zeta(-n) = const1 * Zeta(1+n) if n is odd

and for odd n (i.e. n+1 is even), Zeta(1+n) = const2 * B(1+n)