Chain Reaction

Method 1 : Repeated application of difference of two squares identity , $a^2 -b^2 = (a+b)(a-b)$ .

Multiply the expression by $1 = 2-1$ gives

$\begin{aligned} &&(2-1)(2+1) (2^2 + 1)(2^4 + 1) (2^8 + 1) (2^{16} + 1)(2^{32} + 1)(2^{64} + 1)(2^{128} + 1)(2^{256} + 1)(2^{512} + 1) (2^{1024} + 1) \\ &=& (2^2 - 1)(2^2 + 1)(2^4 + 1) (2^8 + 1) (2^{16} + 1)(2^{32} + 1)(2^{64} + 1)(2^{128} + 1)(2^{256} + 1)(2^{512} + 1) (2^{1024} + 1) \\ &=& (2^4- 1)(2^4 + 1) (2^8 + 1) (2^{16} + 1)(2^{32} + 1)(2^{64} + 1)(2^{128} + 1)(2^{256} + 1)(2^{512} + 1) (2^{1024} + 1) \\ &=& (2^8-1) (2^8+1)(2^{16} + 1)(2^{32} + 1)(2^{64} + 1)(2^{128} + 1)(2^{256} + 1)(2^{512} + 1) (2^{1024} + 1) \\ &=& (2^{16}-1)(2^{16} + 1)(2^{32} + 1)(2^{64} + 1)(2^{128} + 1)(2^{256} + 1)(2^{512} + 1) (2^{1024} + 1) \\ &=& (2^{32} -1)(2^{32} + 1)(2^{64} + 1)(2^{128} + 1)(2^{256} + 1)(2^{512} + 1) (2^{1024} + 1) \\ &=& (2^{64} - 1)(2^{64} + 1)(2^{128} + 1)(2^{256} + 1)(2^{512} + 1) (2^{1024} + 1) \\ &=& (2^{128}- 1)(2^{128} + 1)(2^{256} + 1)(2^{512} + 1) (2^{1024} + 1) \\ &=& (2^{256} - 1)(2^{256} + 1)(2^{512} + 1) (2^{1024} + 1) \\ &=& (2^{512} - 1)(2^{512} + 1) (2^{1024} + 1) \\ &=& (2^{1024} - 1) (2^{1024} + 1) \\ &=& 2^{2048} - 1 \end{aligned}$

Because the base power is minimized, then $a=2, n = 2048,b=1\Rightarrow a+b+n=\boxed{2051}$ .

Method 2 : Find a general pattern.

The expression represents the product of 11 integers.

Let us try to find the pattern of the product of the first few terms (from left to right),

$\begin{aligned} && 2^{2^0} + 1 = 2^1 + 1 = 3 = 2^2 - 1 = 2^{2^1} - 1 \\ && (2^{2^0} + 1)(2^{2^1} + 1) = 3 \times (2^2 + 1) = 3 \times5 = 15 = 2^4 - 1 = 2^{2^2} - 1 \\ && (2^{2^0} + 1)(2^{2^1} + 1)(2^{2^2} + 1) = 15 \times (2^4 + 1) = 15\times17 = 2^8 - 1 = 2^{2^3} - 1 \\ \end{aligned}$

This suggests that the product of the first $n$ terms is equal to $2^{2^n} - 1$ . And this can be proven using induction .

This is obviously true for its base case, $n=0$ gives $2^{2^0} + 1 = 2^{2^1} - 1$ .

Inductive step: If $n=k$ is true for non-negative integer $k$ , then the following equation is true.

$(2^{2^0} + 1)(2^{2^1} + 1)(2^{2^2} + 1) \cdots ( 2^{2^k} + 1)= 2^{2^k} - 1$

Multiplying both sides by $2^{2^{k}} + 1$ gives $(2^{2^0} + 1)(2^{2^1} + 1)(2^{2^2} + 1) \cdots ( 2^{2^k} - 1)(2^{2^{k}} + 1) = 2^{2^{k+1}} - 1$

Which is true as well. Hence our claim is right and so the product of these 11 terms in question is simply $2^{2^{11}} - 1 = 2^{2048} - 1$ .

And because the base power (2) is already minimized, then $a=2,n=2048,b=1\Rightarrow a+b+n=\boxed{2051}$ .