Trivial Zeroes of Riemann Zeta Function are Even Negative Integers

Theorem
The trivial zeroes of the Riemann zeta function occur at the negative even integers, and only there.

More formally, $$\left\{{ s : \zeta(s)=0 \and \left({ \Re(s)< 0 \or \Re(s)> 1 }\right) }\right\} = \left\{{-2,-4,-6,\dots}\right\} \ $$

Proof
We will break the proof into two separate cases:

Case 1: $$\Re(s)>1 \ $$ Case 2: $$\Re(s)<0 \ $$

Case 1
For the region in question,

$$\zeta(s) = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}} \ $$

Therefore, we have

$$\zeta(s) \prod_{p \text{ prime}} \left({ 1-p^{-s} }\right) = 1 \ $$

All of the factors of this infinite product can be found in the product.

$$\prod_{n=2}^\infty \left({ 1-n^{-s} }\right) \ $$

which converges absolutely since the zeta sum $$\sum_{k=1}^\infty k^{-s} \ $$ converges absolutely. Hence

$$\prod_{p \text{ prime}} \left({ 1-p^{-s} }\right) \ $$

converges absolutely and so by the fact that

$$\zeta(s) \prod_{p \text{ prime}} \left({ 1-p^{-s} }\right) = 1 \ $$

we know $$\zeta(s) \ $$ can't possibly be zero for any point in the region in question.

Case 2
For the region in question, the zeta function is defined by the symmetrical relation

$$\Gamma \left({ \frac{s}{2} }\right) \pi^{-s/2} \zeta(s) = \Gamma \left({ \frac{1-s}{2} }\right) \pi^{\tfrac{s-1}{2}} \zeta(1-s) \ $$

Equating $$\zeta(s)=0 \ $$ means

$$0 = \frac{\Gamma \left({ \frac{1-s}{2} }\right) \pi^{\tfrac{s-1}{2}} } {\Gamma \left({ \frac{s}{2} }\right) \pi^{-s/2}}\zeta(1-s)

= \frac{\Gamma \left({ \frac{1-s}{2} }\right) } {\Gamma \left({ \frac{s}{2} }\right) } \pi^{s-\tfrac{1}{2}} \zeta(1-s)

= \frac{\Gamma \left({ \frac{1-s}{2} }\right) } {\Gamma \left({ \frac{s}{2} }\right) }  \zeta(1-s) e^{\left({(s-\tfrac{1}{2}) \log \pi}\right)} $$

by the meaning of powers of $\pi \ $, and since the exponential function is never 0 or infinite, we can divide it out of this expression to get

$$0=\frac{\Gamma \left({ \frac{1-s}{2} }\right) } {\Gamma \left({ \frac{s}{2} }\right) }  \zeta(1-s) \ $$

By Case 1, we know $$\zeta(1-s) \ $$ is not zero in the region in question. By the convergence of the zeta function, we know it is not infinite either. Therefore, we may divide out by this as well to receive

$$0=\frac{\Gamma \left({ \frac{1-s}{2} }\right) } {\Gamma \left({ \frac{s}{2} }\right) } \ $$

By Zeroes of the Gamma Function, we know that $$\Gamma \left({\frac{1-s}{2} }\right) \ $$ is not zero in the region in question either. By Poles of the Gamma Function, we can see that $$\Gamma \left({ \frac{1-s}{2} }\right) \ $$ and $$\Gamma \left({\frac{s}{2} }\right) \ $$ will never both be infinite, and so the danger of a $$\frac{\infty}{\infty} \ $$ situation is avoided and we may divide out by $$\Gamma \left({\frac{1-s}{2} }\right) \ $$ to finally arrive at

$$0 = \frac{1}{\Gamma \left({ \frac{s}{2} }\right) } \ $$

or in other words,

$$\lim_{z\to s} \left|{ \Gamma \left({ \frac{s}{2} }\right) }\right| = \infty \ $$

which we know by Poles of the Gamma Function occurs only at $$s\in \left\{{-2,-4,-6,\dots}\right\} \ $$.