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May I offer an attempt at a very lowbrow answer? This answer might naturally strike anyone who knows the functional equation, who looks at a color coded graph of zeta function such as http://en.wikipedia.org/wiki/File:Complex_zeta.jpg, and who knows a little about the long-range behavior of exponential functions and $\Gamma(z)$. Nothing I will say is beyond the ken of an average serious undergraduate.

$\zeta(s)\rightarrow 1$ as $\Re(s)\rightarrow +\infty$, so no zeros occur far into the right half-plane (the Euler product makes this clear too, indeed no zeros to the right of the critial strip). Then the functional equation prevents also non-trivial zeros far into the left half-plane (indeed to the left of the critical strip).

Now the argument of the function oscillates as you move up vertical lines far into the left half-plane, as one learns from the factors of the functional equation, and their long-range behavior and the right-plane limiting behavior of $\zeta(s)$ itself.

So think about the curves that constitute the real locus and the purely imaginary locus of $\zeta(s)$. These curves can't cross anywhere except on the negative real axis and in the critical strip. Some of these curves do cross on the negative real axis, at the trivial zeros. But there is a rightmost trivial zero, it has its two curves, and there still remain infinitely many other curves lying above those.

No non-trivial zeros would mean no crossing so no doubling back for all these infinitely many curves. What else could they do? They could only head up into the critical strip, packing tightly together. That would imply faster than exponential decay along some locus as you moved up the strip. And why is that impossible?

[I see now that my attempt at a punchline was based on an unjustified assumption. I I'm still thinking about a valid and simple replacement. Suggestions welcome.]

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One can calculate $\zeta(s)$ in the strip by series acceleration: $\zeta(s)$ differs fromthe same series but with alternating signs (actually a Dirichlet L-function) by a simple factor, and then one can split each term of the alternating series into two equal parts and regroup. The (various) periodicities of all the exponential summands means that any behavior we

[I see as we cross the strip horizontally must approximately recur infinitely often higher up in the strip: we just need to find now that my attempt at a punchline was based on an approximate common period for the finitely many terms we consider non-negligible.

Well, if there are no non-trivial zeros then any horizontal trek across the critical strip is bounded away from zero. That makes approximate recurrence incompatible withexponential decay along some locus running up the stripunjustified assumption. I will welcome all suggestions for improvements!!still thinking about a valid and simple replacement. Suggestions welcome.]

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May I offer an attempt at a very lowbrow answer? This answer might naturally strike anyone who knows the functional equation, who looks at a color coded graph of zeta function such as http://en.wikipedia.org/wiki/File:Complex_zeta.jpg, and who knows a little about the long-range behavior of exponential functions and $\Gamma(z)$. Nothing I will say is beyond the ken of an average serious undergraduate.

$\zeta(s)\rightarrow 1$ as $\Re(s)\rightarrow +\infty$, so no zeros occur far into the right half-plane (the Euler product makes this clear too, indeed no zeros to the right of the critial strip). Then the functional equation prevents also non-trivial zeros far into the left half-plane (indeed to the left of the critical strip).

Now the argument of the function oscillates as you move up vertical lines far into the left half-plane, as one learns from the factors of the functional equation, and their long-range behavior and the right-plane limiting behavior of $\zeta(s)$ itself.

So think about the curves that constitute the real locus and the purely imaginary locus of $\zeta(s)$. These curves can't cross anywhere except on the negative real axis and in the critical strip. Some of these curves do cross on the negative real axis, at the trivial zeros. But there is a rightmost trivial zero, it has its two curves, and there still remain infinitely many other curves lying above those.

No non-trivial zeros would mean no crossing so no doubling back for all these infinitely many curves. What else could they do? They could only head up into the critical strip, packing tightly together. That would imply faster than exponential decay along some locus as you moved up the strip. And why is that impossible?

One can calculate $\zeta(s)$ in the strip by series acceleration: $\zeta(s)$ differs from the same series but with alternating signs (actually a Dirichlet L-function) by a simple factor, and then one can split each term of the alternating series into two equal parts and regroup. The (various) periodicities of all the exponential summands means that any behavior we see as we cross the strip horizontally must approximately recur infinitely often higher up in the strip: we just need to find an approximate common period for the finitely many terms we consider non-negligible.

Well, if there are no non-trivial zeros then any horizontal trek across the critical strip is bounded away from zero. That makes approximate recurrence incompatible with exponential decay along some locus running up the strip.

I will welcome all suggestions for improvements!!