The final point $(9)$ is the lemma that I am establishing a proof for, all the relevant lemma are there I guess I just need help fine tuning things, Although I have no idea how I will prove $(10)$ if indeed it is true for all primes.
Consider the Unique prime factorization of a natural number $n$:
$$n=\prod _{j=1}^{\omega \left( n \right) }{p_{{n,j}}}^{\nu_{{n,j}}}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(0)$$
One could, in a heuristic sense, expect that the sum of all sums up to and including the multiplicity of all the distinct prime factors of $n$ will have proportion to the total number of divisors of $n$ somehow.
So we begin by defining the sum of the multiplicities of the unique prime factors of $n$:
$$\Upsilon \left( n \right) =\sum _{j=1}^{\omega \left( n \right) }\nu_{ {n,j}} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(1)$$
$$\sum _{k=1}^{\Upsilon \left( n \right) +1}{\frac { \left( \Upsilon \left( n \right) +1 \right) !}{k!\, \left( \Upsilon \left( n \right) -k+1 \right) !}}=2^{\Upsilon \left( n \right) +1}-1 \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(2)$$
$${\Biggl\{\frac {{2}^{n} \left( {2}^{\Upsilon \left( n \right) +1}-1+{\it \gcd} \left( {2}^{\Upsilon \left( n \right) +1}-1,\tau \left( n \right) \right) \right) }{{\it \gcd} \left( {2}^{\Upsilon \left( n \right) +1},\tau \left( n \right) \right) }} \Biggr\}=0\quad\quad\quad\quad\quad\quad\quad\quad\quad(3)$$
$${\Biggl\{\frac {\tau(n)\left( {2}^{\Upsilon \left( n \right) +1}-1+{\it \gcd} \left( {2}^{\Upsilon \left( n \right) +1}-1,\tau \left( n \right) \right) \right) }{2{\it \gcd} \left( {2}^{\Upsilon \left( n \right) +1},\tau \left( n \right) \right) }} \Biggr\}=0\quad\quad\quad\quad\quad\quad(4)$$
$${\Biggl\{\frac {{2}^{n} }{{\it \gcd} \left( {2}^{\Upsilon \left( n \right) +1},\tau \left( n \right) \right) }} \Biggr\}=0\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(5)$$
$${\Biggl\{\frac {\tau(n)\left( {2}^{\Upsilon \left( n \right) +1}-1 \right) }{2{\it \gcd} \left( {2}^{\Upsilon \left( n \right) +1},\tau \left( n \right) \right) }} \Biggr\}=\frac{1}{2}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(6)$$
$$\exists k \in \mathbb N \land k \gt 1\quad \operatorname{s.t} \sqrt{n}=k \Rightarrow \quad{\Biggl\{\frac {\tau(n)\left({\it \gcd} \left( {2}^{\Upsilon \left( n \right) +1},\tau \left( n \right) \right) \right) }{2{\it \gcd} \left( {2}^{\Upsilon \left( n \right) +1}-1,\tau \left( n \right) \right) }} \Biggr\}=\frac{1}{2}\quad\quad\quad(7)$$
$$\not\exists k \in \mathbb N \land k \gt 1 \quad\operatorname{s.t} \sqrt{n}=k \Rightarrow \quad{\Biggl\{\frac {\tau(n)\left({\it \gcd} \left( {2}^{\Upsilon \left( n \right) +1},\tau \left( n \right) \right) \right) }{2{\it \gcd} \left( {2}^{\Upsilon \left( n \right) +1}-1,\tau \left( n \right) \right) }} \Biggr\}=0\quad\quad\quad(8)$$
$$\exists k \in \mathbb N \land k \gt 1 \quad\operatorname{s.t} \sqrt{n}=k \Rightarrow{\Biggl\{\frac{{\it \gcd} \left( {2}^{\Upsilon \left( n \right) +1},\tau \left( n \right) \right)}{2}}\Biggr\}=\frac{1}{2}\quad\quad\quad(7\operatorname{i})$$
$$\not\exists k \in \mathbb N \land k \gt 1\quad \operatorname{s.t} \sqrt{n}=k \Rightarrow{\Biggl\{\frac{{\it \gcd} \left( {2}^{\Upsilon \left( n \right) +1},\tau \left( n \right) \right)}{2}}\Biggr\}=0\quad\quad\quad(8\operatorname{i})$$
Where ${\{x}\}$ is the fractional part of $x$.
Where $\tau(n)$ is the total number of divisors of $x$.
The 2-adic valuation of the total number of divisors of $n$ can also be expressed in term of the sum of the multiplicities of the unique prime factors of $n$:
$$\nu_{{2}} \left( \tau(n) \right)=\frac{\ln(\gcd(2^{\Upsilon \left( n \right)+1},\tau(n)))}{\ln(2)}\quad\quad\quad(9)$$
The generalization to the p-adic valuation of the total number of divisors of $n$ also seems to be true: $$\nu_{{p}} \left( \tau(n) \right)=\frac{\ln(\gcd(p^{\Upsilon \left( n \right)+1},\tau(n)))}{\ln(p)}\quad\quad\quad(10)$$
Indeed, the following generalizations to any $p$ would appear to be true , giving more substance to the truth value of $(10)$,I am too tired to use my talking words tbh:
$${\Biggl\{\frac {\tau(n)\left( {p}^{\Upsilon \left( n \right) +1}-1+{\it \gcd} \left( {p}^{\Upsilon \left( n \right) +1}-1,\tau \left( n \right) \right) \right) }{{\it \gcd} \left( {p}^{\Upsilon \left( n \right) +1},\tau \left( n \right) \right) }} \Biggr\}=0\quad\quad\quad\quad\quad\quad(4\operatorname{i})$$
$${\Biggl\{\frac {{p}^{n} }{{\it \gcd} \left( {p}^{\Upsilon \left( n \right) +1},\tau \left( n \right) \right) }} \Biggr\}=0\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(5\operatorname{i})$$
$${\Biggl\{\frac {{\it \gcd} \left( {p}^{\Upsilon \left( n \right) +1},\tau \left( n \right) \right) }{{p}^{\Upsilon \left( n \right)-n+1} }} \Biggr\}=0\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(5\operatorname{ii})$$
$$1 \leq {\frac {\tau(n)\left( {p}^{\Upsilon \left( n \right) +1}-1 \right) }{\gcd\left( {p}^{\Upsilon \left( n \right) +1},\tau \left( n \right) \right) }}-p\Biggl\lfloor {\frac {\tau(n)\left( {p}^{\Upsilon \left( n \right) +1}-1 \right) }{p\gcd\left( {p}^{\Upsilon \left( n \right) +1},\tau \left( n \right) \right) }} \Biggr\rfloor \leq p-1\quad\quad\quad\quad\quad\quad(11)$$
Summing up my comments and filling in some gaps:
To show (10) (which includes (9) as a special case):
For any prime $p$ and integer $x$, by definition of the $p$-adic value, we have $\nu_p(x)=\log_p(\gcd(p^r,x))$ if and only if $r$ is an integer $\ge \nu_p(x)$. So equation (10) boils down to the claim that $$\Upsilon(n)+1≥\nu_p(\tau(n))$$ for all $p$ and $n$.
Now to see this, use that $\displaystyle \tau(n) = \prod_{j=1}^{\omega(n)} (\nu_{n,j}+1)$ (according to wikipedia, where $\tau$ is called $\sigma_0$). So for any $p$,
$$\nu_p(\tau(n)) = \sum_{j=1}^{\omega(n)} \nu_p(\nu_{n,j}+1)$$
Now for any integer $a$ and prime $p$, we have $\nu_p(a) \le a-1$, hence we even have the stronger
$$\nu_p(\tau(n)) \le \sum_{j=1}^{\omega(n)} \nu_{n,j} =\Upsilon(n).$$
To show (11):
Calling $x=$ the first fraction in (11), and looking at the function $f(x)=x−p \lfloor x/p\rfloor$, shows that (11) is equivalent to $x$ not being within any of the intervals $(kp−1,kp+1)$ with $k \in \Bbb Z$. Since the denominator of $x$ divides $\tau(n)$, the fraction $x$ is actually an integer itself, so the claim is equivalent to $x$ not being of the form $kp$ with $k \in \Bbb Z$.
This in turn is true by (10), since according to that, the denominator of $x$ is exactly $p^{\nu_p(\tau(n))}$, which exactly cancels the powers of $p$ that are in $\tau(n)$, and the other factor $(p^{whatever}-1)$ is obviously not divisible by $p$.