Let $\; t\in\mathbb{R} \;$ and let $\; \alpha,\beta\in(t,t+1) \;$ with $\; \alpha\not=\beta \;$ and consider $\;\displaystyle\frac{1-\alpha+\beta}{1+\alpha-\beta}$.
My goal was to "simplify" $\; \displaystyle\frac{1-\alpha+\beta}{1+\alpha-\beta} \;$ as the product of terms. In the process of attempting this, I found that $$\frac{1-\alpha+\beta}{1+\alpha-\beta} = \left(1-\alpha+\beta\right)^2 \cdot\prod_{r=1}^\infty\left(1+\left(\alpha-\beta\right)^{2^r}\right)$$
Through a bit of algebraic reworking, I came to the more interesting conclusion that for any $x\in\left[1,\infty\right)$, $$x=\prod_{r=0}^\infty\left(1+\left(\frac{x-1}{x}\right)^{\displaystyle2^r}\right)$$
I'm not sure whether my proof is airtight. I think that (if true) this is a really interesting property. If anyone recognizes it or something similar, please let me know. I tried searching for it but came up with nothing. If anyone would check my proof and/or provide one of their own that would be splendid.
My proof:
$$\frac{1-\alpha+\beta}{1+\alpha-\beta}=\frac{\left(1-\alpha+\beta\right)^2}{\left(1+\alpha-\beta\right) \left(1-\alpha+\beta\right)}=\frac{\left(1-\alpha+\beta\right)^2}{1-\epsilon}$$
where
$$\epsilon=1-\left(1+\alpha-\beta\right)\left(1-\alpha+\beta\right)=\alpha^2+\beta^2-2\alpha\beta=\left(\alpha-\beta\right)^2$$ $$$$
Since $\; \alpha - \beta \in(-1,1)\text{,}\;$ we know that $1> \epsilon >0$.
\begin{align} \frac{\left(1-\alpha+\beta\right)^2}{1-\epsilon}&=\frac{\left(1-\alpha+\beta\right)^2\left(1+\epsilon\right)}{1-\epsilon^2} \\ \\ &= \frac{\left(1-\alpha+\beta\right)^2\left(1+\epsilon\right)\left(1+\epsilon^2\right)}{1-\epsilon^4} \\ \\ &= \frac{\left(1-\alpha+\beta\right)^2\left(1+\epsilon\right)\left(1+\epsilon^2\right)\left(1+\epsilon^4\right)}{1-\epsilon^8} \\ \\ &= \cdots = \frac{\left(1-\alpha+\beta\right)^2\left(1+\epsilon\right)\left(1+\epsilon^2\right)\left(1+\epsilon^4\right)\cdots\left(1+\epsilon^{2^{n-1}}\right)\left(1+\epsilon^{2^n}\right)}{1-\epsilon^{2^{n+1}}}\end{align}
As $n\to\infty$, $\displaystyle\epsilon^{2^{n+1}}\to0$, and we have
\begin{align} \frac{1-\alpha+\beta}{1+\alpha-\beta}&=\left(1-\alpha+\beta\right)^2\left(1+\epsilon\right)\left(1+\epsilon^2\right)\left(1+\epsilon^4\right)\cdots \\ \\ &=\left(1-\alpha+\beta\right)^2 \cdot\prod_{r=0}^\infty\Bigr(1+\epsilon^{2^r}\Bigr) \\ \\ &=\left(1-\alpha+\beta\right)^2 \cdot\prod_{r=0}^\infty \Bigr(1+\left(\left(\alpha-\beta\right)^2\right)^{2^r}\Bigr) \\ \\ &= \left(1-\alpha+\beta\right)^2 \cdot\prod_{r=1}^\infty \Bigr(1+ \left(\alpha-\beta\right)^{2^r}\Bigr) \end{align} $$$$ Dividing both sides of the equation above by $\left(1-\alpha+\beta\right)^2 \text{,}\;$ (a nonzero term), we arrive at
$$\frac{1}{\left(1+\alpha-\beta\right)\left(1-\alpha+\beta\right)} = \frac{1}{\left(1+(\alpha-\beta)\right)\left(1-(\alpha-\beta)\right)} = \prod_{r=1}^\infty \Bigr(1+ \left(\alpha-\beta\right)^{2^r}\Bigr)$$ $$$$
$\alpha - \beta\;$ is an arbitrary value in $\left(-1,1\right) \text{,} \;$ so for any $\gamma\in \left(-1,1\right)$, we have
$$\frac{1}{\left(1+\gamma\right)\left(1-\gamma\right)} = \frac{1}{1-\gamma^2} = \prod_{r=1}^\infty \Bigr(1+ \gamma^{2^r}\Bigr) $$ $$$$
Now let $\;x\in\mathbb{R}\;$ such that $\; x=\frac{1}{1-\gamma^2} \;$ with $\gamma \in \left(-1,1\right)\text{.}\;$ Then $$x=\frac{1}{1-\gamma^2}\iff \gamma=\pm \sqrt{\frac{x-1}{x}}$$ $$$$
We can see here that $\; x \;$ is necessarily an element of $\;\left[1,\infty\right)\;$because $\; x\in\left(-\infty,0\right)\Longrightarrow \gamma>1\;$ (a contradiction), and $\; x\in\left[0,1\right)\Longrightarrow \gamma\not\in\mathbb{R}\;$ (a contradiction). Note that $\; \frac{x-1}{x}\geq 0 \;$ for any $\;x\in\left[1,\infty\right)$.
Thus, if $\; x\in\left[1,\infty\right)\text{,}\;$ then $$x=\prod_{r=1}^\infty\left(1+\left(\sqrt{\frac{x-1}{x}}\right)^{\displaystyle2^r}\right)=\prod_{r=0}^\infty\left(1+\left(\frac{x-1}{x}\right)^{\displaystyle2^r}\right)$$
Example ($x=7$): Wolfram Alpha Example 1
We have a telescoping product,$$\prod_{r\ge0}\left(1+y^{2^r}\right)=\prod_{r\ge0}\frac{1-y^{2^{r+1}}}{1-y^{2^r}}=\frac{1}{1-y}$$for $|y|<1$. Now take $y=1-\frac1x$ for $x\ge1$.