1B. Calculate the product

$$
P=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^{2}}\right)\left(1+\frac{1}{2^{4}}\right)\left(1+\frac{1}{2^{8}}\right) \cdot \ldots \cdot\left(1+\frac{1}{2^{2^{n}}}\right)
$$
Solution. Multiplying both sides of the equation by $1-\frac{1}{2}$ and using the formula for the difference of squares, we get $\left(1-\frac{1}{2}\right) P=1-\frac{1}{2^{2^{n+1}}}$. From this, it follows that $P=2\left(1-\frac{1}{2^{2^{n+1}}}\right)$.