Investigating the q-factiorals. And found experimentally (at least frist 1000 primes) that the continued fraction of $$\frac{\prod_{k=1}^{p-1}(2^k-1)}{2^p-1}$$ for prime p has level 3 and the last element of continued fraction is equal to $p$.
So: $$\frac{\prod_{k=1}^{p-1}(2^k-1)}{2^p-1}=a+\frac{1}{b+\frac{1}{c}}$$ and $c=p$.
If continue: $$\frac{\prod_{k=1}^{p-1}(2^k-1)}{2^p-1}=\frac{cba+a+c}{cb+1}$$
so if $c=p$ $$\frac{\prod_{k=1}^{p-1}(2^k-1)}{2^p-1}=\frac{pba+a+p}{pb+1}$$ Because of Fermat's little theorem, the $b$ is integer. $$2^p-2=pb$$
Now $pb+1=2^p-1$ and $pba+a+p=a(pb+1)+p$ or $pba+a+p=a(2^p-1)+p$ or $$\prod_{k=1}^{p-1}(2^k-1)-p = a(2^p-1)$$.
So if we prove that the $$\prod_{k=1}^{p-1}(2^k-1) \equiv p \pmod {2^p-1}\tag{*}$$
then we prove the initial idea.
There is also very interesting thing about this $$\gcd\big(\prod_{k=1}^{p-1}(2^k-1), {2^p-1}\big)=1$$ This and Fermat's little theorem I assume, should be enough to prove the $(*)$ formula, but hang on the proving $(*)$ .
Let $\omega \equiv2 \bmod{(2^p-1)}$, where we have $\omega^p = 1$. It follows that: $$\prod_{k=1}^{p-1} (2^k-x) \equiv \prod_{k=1}^{p-1} (\omega^k-x) = 1+x+\cdots+x^{p-1}$$ and substituting $x = 1$ yields the product being $p$ modulo $2^p - 1$. Now, it follows that: $$\frac{\prod_{k=1}^{p-1} (2^k-1)}{2^p-1} = a + \frac{p}{2^p-1} = a + \frac{1}{b + \frac{1}{p}}$$ where $b = \frac{2^p-2}{p}$ is a positive integer by Fermat's Little Theorem.
Clarification: The use of $\omega$ is to provide intuition that we our arithmetic matches with that of $p$th roots of unity. It is possible to formalize this notion.
In the field $\mathbb{Q}$, we know that $y^{p-1}+y^{p-2}+\cdots + 1$ is an irreducible polynomial in $\mathbb{Q}[y]$. Now, we consider the field extension $\mathbb{Q}[y]/(y^{p-1}+\cdots + y + 1)$. We have the roots of the polynomial $x^{p-1}+x^{p-2} + \cdots + 1$ are the roots of $x^p = 1$ which are not equal to $1$. Hence, they are given by $y, y^2, \ldots, y^{p-1}$. It follows that: $$\prod_{k=1}^{p-1} (x - y^k) = x^{p-1}+x^{p-2}+\cdots+1$$ However, this means that the identity is in fact true in $\mathbb{Z}[x,y]/(y^{p-1}+y^{p-2}+\cdots+1)$. It is then also true in $(\mathbb{Z}/(2^p-1))[x, y]/(y^{p-1}+y^{p-2}+\cdots + 1)$. We can substitute $x = 1$ and $y = 2$ to yield the required. Note that we may substitute $y = 2$ since it satisfies $y^{p-1}+y^{p-2}+\cdots + 1 = 0$ in $\mathbb{Z}/(2^p-1)$.