Why doesn't the Riemann Zeta Function have a zero at $s=0$?

Question

I have a few questions regarding the Riemann Zeta Function and its zeros. First, I will state what is clear to me. I see that $\zeta (s)$ is clearly defined and nonzero when $Re(s)>1$ since then $\zeta (s)$ is defined by convergent power series. I know that $\zeta (s)$ has a pole at $s=1$ of residue $1$. Likewise, I am familiar with the function equation $\pi ^{-\frac{s}{2}}\Gamma (\frac{s}{2})\zeta(s)=\pi ^{-\frac{1-s}{2}}\Gamma (\frac{1-s}{2})\zeta (1-s)$. From this equation, I see that $-2,-4,\dots $ are zeros, since $\Gamma$ has poles there. I also know that we refer to the the nontrivial zeros as those in the critical strip, i.e. $0<Re(s)<1$. Here are my questions:

1. How do we know that there are no nontrivial zeros to the left of the critical strip, i.e. when $Re(s)<0$?
2. How do we know that there are no zeros where $Re(s)=0$?
3. Why is $s=0$ not a zero of $\zeta (s)$? This question is particularly key to me, since $\Gamma(s)$ is not defined at any nonpositive integer. By the same reasoning we used to deduce that the negative even integers are zeros using the functional equation above, shouldn't $0$ also be a zero? Since $\Gamma(s)$ is not defined, I don't see how this doesn't violate the functional equation.
4. How do we know that $\zeta (s)$ has no zeros on the line $Re(s)=1$?

I understand that I am literally asking multiple questions, but I am hoping that a single clarification will help me answer all four.

Answer 1

1. If you know that $\Gamma(s)$ doesn't have zeros (i.e. $1/\Gamma(s)$ is an entire function) and that $\zeta(s)$ doesn't have zeros in $\Re s>1$ (because it is represented by Euler product there), then the answer follows from the functional equation: $\Gamma(s/2)\zeta(s)$ doesn't have zeros in $\Re s<0$.
2. This follows from the functional equation and item 4.
3. Again, follows from the functional equation (if $s=0$ were a zero of $\zeta(s)$ then $\Gamma(s/2)\zeta(s)$ would stay regular at $s=0$).
4. This is the core of an existing proof of PNT, see the end of this section.

Answer 2

3. Yes, $\Gamma(\frac{s}{2})$ has a pole at $s=0$. However, calculating $\zeta(s)$ at $s=0$ cannot be settled with that argument alone due to the fact that $\zeta(1-s)$ also has a pole at $s=0$, meaning that you'd need to resolve an $\frac{\infty}{\infty}$ situation to calculate the actual value of $\zeta(s)$.

Answer 3

1. Consider the functional equation you give, along with some general knowledge of the $\Gamma$ function (it has no roots, and only poles at non-positive integers). No roots for $Re(s)>1$ then implies no roots for $Re(s)<0$, apart from the trivial ones.