Powerful Product

Notice that $2^{2^{n+1}}=2^{2^n\times2}=(2^{2^n})^2$ , hence we can rewrite the never-ending apple pie above as $\begin{aligned} \large\prod _{n=0}^\infty\left[\left(\frac1{2^{2^{n}}}\right)^2-\frac1{2^{2^n}}+1\right] &=\large\prod _{n=0}^\infty\frac{\left(\frac1{2^{2^{n}}}\right)^3+1}{\frac{1}{2^{2^n}}+1}\\ &=\frac{\left[\left(\frac1{2^{2^{0}}}\right)^3+1\right]\left[\left(\frac1{2^{2^{1}}}\right)^3+1\right]\left[\left(\frac1{2^{2^{2}}}\right)^3+1\right]\ldots}{\left[\frac{1}{2^{2^0}}+1\right]\left[\frac{1}{2^{2^1}}+1\right]\left[\frac{1}{2^{2^2}}+1\right]\ldots} \end{aligned}$

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Consider the numerator, if we multiply it by $\left(\frac1{2^{2^{0}}}\right)^3-1$ , then we have $\begin{aligned} M&=\left[\left(\frac1{2^{2^{0}}}\right)^3-1\right]\left[\left(\frac1{2^{2^{0}}}\right)^3+1\right]\left[\left(\frac1{2^{2^{1}}}\right)^3+1\right]\left[\left(\frac1{2^{2^{2}}}\right)^3+1\right]\ldots \\&=\left[\left(\frac1{2^{2^{0}}}\right)^6-1^2\right]\left[\left(\frac1{2^{2^{1}}}\right)^3+1\right]\left[\left(\frac1{2^{2^{2}}}\right)^3+1\right]\ldots \\&=\left[\left(\frac1{2^{2^{1}}}\right)^3-1\right]\left[\left(\frac1{2^{2^{1}}}\right)^3+1\right]\left[\left(\frac1{2^{2^{2}}}\right)^3+1\right]\ldots \\&=\left[\left(\frac1{2^{2^{2}}}\right)^3-1\right]\left[\left(\frac1{2^{2^{2}}}\right)^3+1\right]\ldots \\&=\ldots \end{aligned}$ Observe that we are just repeatedly applying the identity $(a-b)(a+b)=a^2-b^2$ , thus $\begin{aligned} M&=\lim_{k\to\infty}\left\{\left[\left(\frac1{2^{2^{0}}}\right)^3-1\right]\large\prod_{n=0}^k\left[\left(\frac1{2^{2^{n}}}\right)^3+1\right]\right\} \\&=\lim_{k\to\infty}\left[\left(\frac{1}{2^{2^{k+1}}}\right)^3-1\right] \\&=-1 \end{aligned}$

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Similarly, consider the denominator, if we multiply it by $\frac1{2^{2^{0}}}-1$ , then we have $\begin{aligned} N&=\left(\frac1{2^{2^{0}}}-1\right)\left(\frac1{2^{2^{0}}}+1\right)\left(\frac1{2^{2^{1}}}+1\right)\left(\frac1{2^{2^{2}}}+1\right)\ldots \\&=\left(\frac1{2^{2^{1}}}-1\right)\left(\frac1{2^{2^{1}}}+1\right)\left(\frac1{2^{2^{2}}}+1\right)\ldots \\&=\left(\frac1{2^{2^{2}}}-1\right)\left(\frac1{2^{2^{2}}}+1\right)\ldots \\&=\ldots \end{aligned}$ Again we are just repeatedly applying the identity $(a-b)(a+b)=a^2-b^2$ , thus $\begin{aligned} N&=\lim_{k\to\infty}\left[\left(\frac1{2^{2^{0}}}-1\right)\large\prod_{n=0}^k\left(\frac1{2^{2^{n}}}+1\right)\right] \\&=\lim_{k\to\infty}\left(\frac{1}{2^{2^{k+1}}}-1\right) \\&=-1 \end{aligned}$

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Now... $\begin{aligned} \large\prod _{n=0}^\infty\left(\frac1{2^{2^{n+1}}}-\frac1{2^{2^n}}+1\right)&=\frac{\left[\left(\frac1{2^{2^{0}}}\right)^3+1\right]\left[\left(\frac1{2^{2^{1}}}\right)^3+1\right]\left[\left(\frac1{2^{2^{2}}}\right)^3+1\right]\ldots}{\left[\frac{1}{2^{2^0}}+1\right]\left[\frac{1}{2^{2^1}}+1\right]\left[\frac{1}{2^{2^2}}+1\right]\ldots} \\&=\frac{\left(\frac1{2^{2^{0}}}-1\right)M}{\left[\left(\frac1{2^{2^{0}}}\right)^3-1\right]N} \\&=\frac{1-\frac{1}{2}}{1-\frac{1}{8}} \\&=\frac{\frac12}{\frac78} \\&=\frac47 \end{aligned}$ $\therefore A+B=4+7=11$

Phew, that was a lot of typing, imma gonna drink ma cup o tea now :P

Hint: Use the fact that $\dfrac{x^3+1}{x+1}=x^2-x+1$ . (Will write up a more complete solution later)

The infinite product is $\dfrac47$ .