= Solution
For $\operatorname{Re}s>1$, define the <Riemann zeta function> by the absolutely convergent <Dirichlet series>
$$
\zeta(s)=\sum_{n\geq1}n^{-s}=\prod_p(1-p^{-s})^{-1}.
$$
The <Euler product> follows from <unique prime factorization> and <absolute convergence>. The counting-function integral gives
$$
\zeta(s)=s\int_1^\infty\lfloor u\rfloor u^{-s-1}\,du
=\frac{s}{s-1}-s\int_1^\infty\{u\}u^{-s-1}\,du.
$$
Because the <fractional part> is bounded, the last integral, and its derivatives with respect to $s$ on compact subsets, converge uniformly for $\operatorname{Re}s>0$. It is a <holomorphic function> there. This proves the <Meromorphic continuation of the Riemann zeta function to the right half-plane>, with a <simple pole> at one of <residue> one and no other singularity in that half-plane.
The <Gamma function> is
$$
\Gamma(z)=\int_0^\infty e^{-u}u^{z-1}\,du\qquad(\operatorname{Re}z>0).
$$
The <Gamma function recurrence> extends it meromorphically; its poles are at the nonpositive integers and it has no zeros. The <functional equation of the Riemann zeta function> can be stated as
$$
\boxed{\pi^{-s/2}\Gamma(s/2)\zeta(s)
=\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s),}
$$
or, equivalently,
$$
\boxed{\zeta(s)=2^s\pi^{s-1}\sin(\pi s/2)\Gamma(1-s)\zeta(1-s).}
$$
These are identities of <meromorphic functions>; the second also extends the <Riemann zeta function> to the remaining half-plane. The <critical strip> is $0<\operatorname{Re}s<1$; its boundary lines will also be treated below, rather than included among possible exceptional zeros.
For $\operatorname{Re}s>1$, the absolutely convergent reciprocal <Euler product> gives $\zeta(s)\neq0$. For $\operatorname{Re}s<0$, the factors $2^s$, $\pi^{s-1}$, $\Gamma(1-s)$ and $\zeta(1-s)$ in the second <functional equation of the Riemann zeta function> are finite and nonzero. Hence the only zeros there are the zeros of the sine factor:
$$
\boxed{s=-2,-4,-6,\ldots.}
$$
Each is simple. At zero, the sine zero cancels the pole of $\zeta(1-s)$; using its <residue> one gives $\zeta(0)=-1/2$, not zero. Nonvanishing on the rest of $\operatorname{Re}s=0$ follows from the boundary-line proof below and the <functional equation of the Riemann zeta function>. Together these facts prove that the <trivial zeros of the Riemann zeta function> are the only zeros outside the open <critical strip>.
Here is a <Jensen disk proof of the zeta zero-count bound> which avoids any unproved left-half-plane growth estimate. Set $F(s)=(s-1)\zeta(s)$, a <holomorphic function> throughout $\operatorname{Re}s>0$, including at one. The integral continuation formula gives
$$
|F(s)|\ll(|s|+1)^2\qquad(\operatorname{Re}s\geq1/4).
$$
For each integer $j$, apply <Jensen's formula> in the disk with centre $2+ij$, outer radius $R=7/4$ and inner radius $r=8/5$. The outer disk stays in $\operatorname{Re}s\geq1/4$, so its maximum modulus is $O((|j|+2)^2)$. At its centre,
$$
|\zeta(2+ij)|\geq\zeta(2)^{-1},
\qquad |F(2+ij)|\geq|1+ij|/\zeta(2),
$$
since $|1/\zeta(2+ij)|\leq\sum_n|\mu(n)|n^{-2}\leq\zeta(2)$. Therefore the number of zeros in its inner disk, counted with multiplicity, is at most
$$
\frac{\log\bigl(\max_{|s-(2+ij)|\leq R}|F(s)|/|F(2+ij)|\bigr)}{\log(R/r)}
\ll\log(|j|+2).
$$
The rectangle $1/2\leq\operatorname{Re}s\leq1$, $|\operatorname{Im}s-j|\leq1/2$ fits in the inner disk because its farthest point has distance $\sqrt{(3/2)^2+(1/2)^2}=\sqrt{5/2}<8/5$. Summing over $O(T)$ such disks counts $O(T\log T)$ zeros in the right half of the <critical strip> up to height $T$. The first <functional equation of the Riemann zeta function> bijects zeros, with multiplicities, in the left half with their reflections $s\mapsto1-s$ in the right half; its <Gamma function> factors are finite and nonzero in the strip. Thus
$$
\boxed{\#\{\rho:\zeta(\rho)=0,\ 0<\operatorname{Re}\rho<1,\ |\operatorname{Im}\rho|\leq T\}\ll T\log T.}
$$
The pole of <zeta function> at one does not count as a zero: $F(1)=1$.
Finally, for real $\sigma>1$ and $t\neq0$, the Euler-product logarithms and the nonnegative <trigonometric polynomial>
$$
3+4\cos\theta+\cos(2\theta)=2(1+\cos\theta)^2\geq0
$$
give the <three-four-one product proof of zeta boundary nonvanishing>:
$$
\log\bigl(\zeta(\sigma)^3|\zeta(\sigma+it)|^4|\zeta(\sigma+2it)|\bigr)
=\sum_{p,k\geq1}\frac{3+4\cos(kt\log p)+\cos(2kt\log p)}{kp^{k\sigma}}\geq0.
$$
If $\zeta(1+it)$ had a zero of order $a\geq1$, its factor would be $O((\sigma-1)^{4a})$ as $\sigma\downarrow1$. The real <zeta function> factor has pole order three, and the factor at $1+2it$ stays bounded because $t\neq0$. The product would tend to zero as $O((\sigma-1)^{4a-3})$, contradicting that it is at least one. Therefore
$$
\boxed{\zeta(1+it)\neq0\qquad(t\neq0).}
$$
At $t=0$ there is a pole, not a zero. Applying the sine-form <functional equation of the Riemann zeta function> at $s=it\neq0$ now proves nonvanishing on the remaining imaginary axis, completing the earlier assertion about all zeros outside the <critical strip>.
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