Closed form for $(2^1-1)(2^2-1)...(2^k-1)$?

Question

Is there closed form for $\prod_1^{i=k}(2^i-1)$ ?

I found that it is the product of the terms of the following arithmetico-geometric sequence : $$\{u_1=1,u_{n+1}=2u_n+1\}$$ I found nothing with factorials, but there may be something.

For the story, I want to prove that : $$\forall n \in \Bbb N, n\gt 3, \exists m \in \Bbb N, 0 \lt m \lt n$$ such as $$\forall k \in \Bbb N, 0 \lt k \lt n^2-m²-2n+1, 2^k \not\equiv 1 \mod (n^2-m^2)$$ and $$2^{n^2-m^2-2n+1} \equiv 1 \mod (n^2-m^2)$$ That's why I tried to make the product.

Answer 1

In terms of known notations:

Let $u_{n+1} = 2 \, u_{n} + 1$ where $u_{0}=1$, for which $u_{n} = 2^{n}-1$. Using $$(x;q)_{n} = \prod_{k=0}^{n-1} (1 - x \, q^{r})$$ leads to \begin{align} P_{n} &= \prod_{k=0}^{n-1} \{ u_{k} \} = \prod_{k=1}^{n} \{ 2^{k+1} - 1 \} \\ &= 2^{\binom{n+1}{2}} \, \prod_{k=1}^{n} \left(1 - \frac{1}{2^{k}}\right) \\ &= 2^{\binom{n+1}{2}} \, \left(\frac{1}{2}; \frac{1}{2}\right)_{n}. \end{align}

Answer 2

The (Gauss) $q$-analogue of the binomial theorem expresses the product of terms $(1 - q^i)$ as a sum whose terms are (up to some normalization factors) $q$-binomial coefficients.

Formulas are listed at Gaussian binomial coefficients in Wikipedia, for example.