$$ \prod_{k=0}^{n} \frac{\left( \left( \cot x \right)^{2^k} + 1 \right)^2}{\left( \left( \cot x \right)^{2^{k+1}} + 1 \right)}$$
My approach
I tried to convert the entire expression in terms of tan(x) and that led to essentially the same expression with tan instead of cot. From there I tried to open the square up and break it into two fractions. However this didn't yield me a telescopic series of any sort. I also thought of approaching the question with limit of a sum but couldn't get through.
This is a revised answer found with the assistance of user326159. My original answer is below the horizontal line.
\begin{align*}\prod_{k=0}^{n} \frac{(\cot^{2^k}(x) + 1)^2}{\cot^{2^{k+1}}(x) + 1 } & = \frac{(\cot(x) + 1)^2}{\cot^{2}(x) + 1 }\frac{(\cot^{2}(x) + 1)^2}{\cot^{4}(x) + 1}\frac{(\cot^4(x) + 1)^2}{\cot^{8}(x) + 1 } \cdots \frac{\left(\cot^{2^n}(x) + 1 \right)^2}{\cot^{2^{n+1}}(x) + 1 } \\ & = (\cot(x) + 1)^2(\cot^{2}(x) + 1)(\cot^{4}(x) + 1)(\cot^{8}(x) + 1) \cdots \frac{\cot^{2^n}(x) + 1}{\cot^{2^{n+1}}(x) + 1 }\\ & =\frac{\cot(x) + 1}{\cot^{2^{n+1}}(x) + 1 }\prod_{k=0}^{n} \left(\cot^{2^k}(x) + 1 \right). \end{align*}
The remaining part of the answer uses difference-between-two-squares factorization:
$$(f(x))^2-1 = (f(x)+1)(f(x)-1)$$
Therefore:
$$f(x)+1 = \frac{(f(x))^2 - 1}{f(x) - 1 }$$
From this result:
$$\cot^{2^k} (x)+1 = \frac{\cot^{2^{k+1}}(x) - 1}{\cot^{2^k}(x)-1 }$$
Therefore:
\begin{align*}\prod_{k=0}^{n} \frac{(\cot^{2^k}(x) + 1)^2}{\cot^{2^{k+1}}(x) + 1 } & = \frac{\cot(x) + 1}{\cot^{2^{n+1}}(x) + 1 }\prod_{k=0}^{n} \frac{\cot^{2^{k+1}} (x) - 1}{\cot^{2^k} (x)-1 }\\ & = \frac{\cot(x) + 1}{\cot^{2^{n+1}}(x) + 1 }\left(\frac{\cot^{2}(x) - 1}{\cot(x) - 1 }\frac{\cot^{4}(x) - 1}{\cot^{2}(x) - 1}\frac{\cot^{8}(x) - 1}{\cot^{4}(x) - 1} \cdots \frac{\cot^{2^{n+1}}(x) - 1}{\cot^{2^{n}}(x) - 1}\right) \\ & = \frac{(\cot(x) + 1) \left (\cot^{2^{n+1}}(x) - 1 \right )}{(\cot(x) - 1) \left (\cot^{2^{n+1}}(x) + 1 \right ) } \end{align*}
The difference-between-two-squares factorization probably could have been used from the start to get the same result in fewer steps, but I think there would be messier intermediate algebra.
For this product, the numerator when k=i+1 is equal to the denominator squared when k=i. This reduces the product to the numerator at k=0 multiplied by the denominator at k=n multiplied by a simpler product. The simpler product is the expression within the outer parentheses of the numerator from k=1 to n-1.
I don't think that simpler product yields a value, but it can be written as an expression using a q-Pochhammer symbol.
My guess is that this problem was once part of a larger problem that gave a value for x.
\begin{align*}\prod_{k=0}^{n} \frac{\left[\cot^{2^k}(x) + 1 \right]^2}{\cot^{2^{k+1}}(x) + 1 } & = \frac{\left[\cot(x) + 1 \right]^2}{\cot^{2}(x) + 1 }\frac{\left[\cot²(x) + 1 \right]^2}{\cot^{4}(x) + 1}\frac{\left[\cot^4(x) + 1 \right]^2}{\cot^{8}(x) + 1 } \cdots \frac{\left[\cot^{2^n}(x) + 1 \right]^2}{\cot^{2^{n+1}}(x) + 1 } \\ & = \frac{1}{\cot^{2^{n+1}}(x) + 1 }\prod_{k=0}^{n} \left[\cot^{2^k}(x) + 1 \right] \\ & = \frac{1}{\cot^{2^{n+1}}(x) + 1 }\sum_{k=0}^{2^n-1} \cot^{k}(x) \end{align*}
