Power of 2 product

Given that

$\begin{aligned} \prod_{k=1}^{\blue{\log_2 n}} \left(x^{\frac {\blue n}{2^k}} + 1\right) & = \prod_{k=1}^{\blue m} \left(x^{\frac {\blue{2^m}}{2^k}} + 1\right) & \small \blue{\text{Let }m = \log_2 n \implies n = 2^m} \\ & = \prod_{k=1}^{m} \left(x^{2^{m-k}} + 1\right) \\ & = \left(x^{2^{m-1}}+1\right) \left(x^{2^{m-2}}+1\right) \left(x^{2^{m-3}}+1\right) \cdots \left(x+1\right) \\ & = \left(x+1\right) \left(x^2+1\right) \left(x^4+1\right) \cdots \left(x^{2^{m-1}}+1\right) \\ & = \frac {\blue{(x-1)} \left(x+1\right) \left(x^2+1\right) \left(x^4+1\right) \cdots \left(x^{2^{m-1}}+1\right)}{\blue{x-1}} \\ & = \frac {\left(x^2-1\right) \left(x^2+1\right) \left(x^4+1\right) \left(x^8+1\right) \cdots \left(x^{2^{m-1}}+1\right)}{x-1} \\ & = \frac {\left(x^4-1\right) \left(x^4+1\right) \left(x^8+1\right) \left(x^{16}+1\right) \cdots \left(x^{2^{m-1}}+1\right)}{x-1} \\ & = \quad \cdots \quad \cdots \quad \cdots \\ & = \frac {x^{2^m}-1}{x-1} = \frac {x^n-1}{x-1} \end{aligned}$

Therefore, $A=n$ and $B = C = 1$ . For $n=2^{92} \implies \log_2 A + B + C = 92+1+1 = \boxed{94}$ .

$\dfrac{x^{n} -1}{x-1}$ can be written as $\dfrac{\left(x^{\frac{n}{2}} -1 \right) \left(x^{\frac{n}{2}} +1 \right)}{x-1}$

if n is a power of 2 and we keep continuing this process eventually the (x-1) in denominator will cancel leaving a product of the form in the question. the number of times this iteration happens depends on $\log_{2} n$ . therefore the product in question is = $\dfrac{x^{n} -1}{x-1}$ Answer becomes 92+1+1=94

This product is also the sum of geometric series with ratio x from 1 to $x^{n-1}$