Note. I have added some remarks at the end of my post regarding the mistakes in my argument.
The other day I proved that $\zeta(s)^2=\sum_{n=1}^\infty\nu(n)n^{-s}$ for all $s\in\mathbb R,$ where $\nu(n)$ is the number of positive divisors of $n,$ however I'm not sure about my argument, which is the following (I'm sorry if I'm not seeing some trivial detail).
Proposition. Prove that if $s$ is a real number then $$\sum_{n=1}^\infty\dfrac{\nu(n)}{n^s}=\zeta(s)^2.$$
Proof. If $s\leqslant1$ the result is clear. Now assume $s>1.$ For each positive integer $n$ define $$f(n):=\sum_{d\mid n}\dfrac{\nu(d)}{d^s},$$ where the sum runs over all positive divisors of $n.$ Note that since the function $\nu(n)/n^s$ is multiplicative then $f$ is multiplicative; thus for a fixed $n$ we have $$f\left(\prod_{p\leqslant n}p^{\alpha_p}\right)=\prod_{p\leqslant n}f(p^{\alpha_p}),\tag1$$ where the products are taken over all positive primes not exceeding $n$ and where $\alpha_p:=\lfloor\log{n}/\log p\rfloor$ for each $p.$ Now fix $p.$ Clearly we have $$f(p^{\alpha_p})=\sum_{m=0}^{\alpha_p}\dfrac{m+1}{p^{ms}},$$ so if we set $t:=\alpha_p,$ $q:=p^s$ and $\sigma:=1+q^{-1}+\cdots+q^{-t}$ then we have $$\begin{aligned} f(p^t)&=\sigma+(\sigma-1)+(\sigma-1-q^{-1})+\cdots+(\sigma-1-q^{-1}-\cdots-q^{1-t})\\\\&=(t+1)\sigma-t-(t-1)q^{-1}-\cdots-(t-(t-1))q^{1-t}\\\\&=(t+1)\sigma-t(\sigma-q^{-t})+q^{-1}\left(1+2q^{-2}+\cdots+(t-1)^{q^{2-t}}\right)\\\\&=\sigma+tq^{-t}+q^{-1}\left(f(p^t)-tq^{-t}-(t+1)q^{-t}\right)\end{aligned}$$ and hence $$f(p^t)(1-q^{-1})^2=1+tq^{-t}+2tq^{-t-2}-2q^{-t-1}-3tq^{-t-1}-q^{-t-2}$$ but if $p$ is fixed and we let $n\to\infty$ then each of the terms to the right of $1$ in the last expression goes to $0$ (as a consequence of the binomial theorem) so $f(p^t)\to(1-q^{-1})^{-2}=(1-p^{-s})^{-2}$ which by $(1)$ implies that $$\lim_{n\to\infty}f\left(\prod_{p\leqslant n}p^{\alpha_p}\right)=\prod_p(1-p^{-s})^{-2}=\zeta(s)^2\tag2$$ and this together with $$\sum_{k\leqslant n}\dfrac{\nu(k)}{k^s}\leqslant f\left(\prod_{p\leqslant n}p^{\alpha_p}\right)=\sum_{d\mid\prod\limits_{p\leqslant n}p^{\alpha_p}}\dfrac{\nu(d)}{d^s}<\sum_{k=1}^\infty\dfrac{\nu(k)}{k^s}$$ implies that $$\sum_{n=1}^\infty\dfrac{\nu(n)}{n^s}=\zeta(s)^2,$$ as desired. $\square$
THE PROBLEM IS is it right to conclude that if $n\to\infty$ then $\alpha_p\to\infty$ for each prime $p$?
EDIT. A few minutes ago (like two hours ago) I realized what was wrong in my proof. None of the answers I received was helpful in that particular sense (althoug they were helpful). First, let me clarify why I say that the case $s\leqslant1$ trivial: I assume (maybe my assumption is completely incorrect) that $\zeta(s)^2=(\lim_{n\to\infty}\sum_{k=1}^n k^{-s})^2.$ Now let me clarify where is the main mistake in my argument: of course for a fixed prime $p$ we have $\alpha_p=\lfloor\log{n}/\log{p}\rfloor\to\infty$ as $n\to\infty,$ however, this doesn't necessarily implie that $\lim_{n\to\infty}f\left(\prod_{p\leqslant n}p^{\alpha_p}\right)=\prod_p(1-p^{-s})^{-2}.$ As an easy example of a case in which this same argument can be (wrongly) applied is in the sequence $\{\sqrt[n]{n!}\}.$ We could, using the same (wrong) reasoning I used and fix $m,$ with $1\leqslant m\leqslant n.$ Then we observe that $\sqrt[n]{m}\to1$ as $n\to\infty$ and conclude (wrongly) that therefore $\sqrt[n]{n!}\to1.$
For $s,t\in\mathbb{R}$ and $s>\max\{1,r+1\}$ we have
$$\zeta(s)\zeta(s-r)=\sum\limits_{t=1}^\infty\frac{1}{t^s}\sum\limits_{d=1}^\infty\frac{1}{d^{s-r}}=\sum\limits_{n=1}^\infty\frac{1}{n^s}\sum\limits_{t\cdot d=n}d^r$$
and with $r:=0$ we get the formula for $\zeta^2(s)$.