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There are infinitely many prime numbers. Indeed, for any finite list of prime numbers $p_1,\dots ,p_k$, the integer $N=p_1\cdots p_k+1>1$ has a prime factor, by the Fundamental theorem of arithmetic. None of the listed prime numbers divides $N$, since each divides $N-1$ and a common divisor would divide $1$. Thus no finite list contains all prime numbers. The constructed $N$ need not itself be a prime number.

Absolute convergence and the Fundamental theorem of arithmetic give the Euler productFor a finite set of primes, expand the geometric factors: their product sums $n^{-s}$ over integers whose prime factors lie in that set. Let the finite sets increase through all primes. Absolute convergence permits passage to the limit and recovers the full Dirichlet series. Moreover ${\sum }_{p,m\ge 1}|p^{-ms}|/m<\infty$, so the logarithm converges and the product has no zeros there.

The same absolutely convergent logarithm, and $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2\ge 0$, giveThis is the product version of the three-four-one zero-free-region argument. It also proves there are no zeros on $\Re s=1$: if $\zeta (1+i{t}_0)=0$ for $t_0\ne 0$, its factor has order at least four as $\sigma \downarrow 1$, while the real pole contributes only order minus three and the $2t_0$ factor remains bounded. The displayed left side would tend to zero, a contradiction. At $t_0=0$ there is a pole, not a zero.

For large $|t|$, put $L=\log (|t|+2)$. The Hardy-Littlewood approximation to the Riemann zeta function at $x\asymp |t|$ gives $|\zeta (\sigma +it)|\ll L$ for $1-2/L\le \sigma \le 3$; its finite sum is bounded by ${\sum }_{n\le x}{n}^{-1+2/L}\ll L$ and the integral term is bounded. The Cauchy estimate for derivatives on circles of radius comparable to $1/L$ consequently gives $|{\zeta }^{\prime }(\sigma +it)|\ll L^2$ for $1-L^{-9}\le \sigma \le 2$.

Take ${\sigma }_0=1+a{L}^{-9}$ with a small fixed $a>0$. The product inequality, $\zeta ({\sigma }_0)\ll L^9/a$, and $|\zeta ({\sigma }_0+2it)|\ll L$ implyIf $1-c{L}^{-9}\le \sigma \le {\sigma }_0$, integration of the derivative along the horizontal segment changes this value by at most $C(a+c)L^{-7}$. Choose $a$ sufficiently small that $Ca<c_0{a}^{3/4}/4$, then $c\le a$ sufficiently small. The lower bound remains a positive multiple of $L^{-7}$. For ${\sigma }_0\le \sigma \le 2$, the same product inequality, $\zeta (\sigma )\le \zeta ({\sigma }_0)$, and the near-one upper bound give that lower bound directly. For $\sigma \ge 2$, the reciprocal Euler product gives $|1/\zeta (s)|\le \zeta (2)$.

Finally the no-zero result on $\Re s=1$, compactness at bounded heights and the regular reciprocal at the pole allow a further fixed reduction of $c$ to include bounded $t$. We have proved the weak logarithmic zero-free region for the Riemann zeta functionThe reciprocal at $s=1$ is its holomorphic extension, equal to zero.

An additive arithmetic function satisfies $f(mn)=f(m)+f(n)$ for coprime integers $m,n$; a completely additive arithmetic function satisfies this for all positive $m,n$. In either case $f(1)=0$. The Fundamental theorem of arithmetic gives $f(n)={\sum }_{p^k\parallel n}f(p^k)$, where $p^k\parallel n$ means that the prime valuation of $n$ at $p$ is exactly $k$.

Under the discrete uniform distribution on $[N]$, the expectation of the indicator function of $p^k\parallel n$ isThe formula remains valid when $p^{k+1}>N$. Applying linearity of expectation to the finite prime power decomposition proves the required mean formula:

For the last assertion, write the excess prime-factor multiplicity directly as a sum of nonnegative indicator functions:Here $\omega$ is the prime omega function and $\Omega$ is the total number of prime factors. The expectation is bounded independently of $N$, sinceThus Markov inequality gives vanishing probability on every diverging scale: