Raise to the Raise

By using a difference of two squares we can show that: $\large 2^{2^i} +1 = \frac{2^{2^{i+1}} -1}{2^{2^{i}} -1}$ This creates a very neat telescoping series: $\large \displaystyle\prod_{i=o}^{1024} ( 2^{2^i} +1) = \displaystyle\prod_{i=o}^{1024} (\frac{2^{2^{i+1}} -1}{2^{2^{i}} -1})$ $\large = \frac{2^2-1}{2-1}\times\frac{2^4-1}{2^2-1}\times\ ... \times\frac{2^{2^{1024}} -1}{2^{2^{1023}}-1}\times\frac{2^{2^{1025}} -1}{2^{2^{1024}}-1}$ $\Huge = 2^{2^{1025}} - 1 = \boxed{2^{2^{(2^{10} +1)}}}$

$We\quad can\quad write\quad the\quad value\quad as-\\ \\ \left( 2+1 \right) \left( 4+1 \right) \left( 16+1 \right) \left( 256+1 \right) .....\left( { 2 }^{ { 2 }^{ 1023 } }+1 \right) \left( { 2 }^{ { 2 }^{ 1024 } }+1 \right) \\ \\ Since\quad (2-1)\quad is\quad 1,\quad we\quad can\quad multiply\quad it\quad to\quad anything.\\ \\ \left( 2+1 \right) \left( 4+1 \right) \left( 16+1 \right) \left( 256+1 \right) .....\left( { 2 }^{ { 2 }^{ 1023 } }+1 \right) \left( { 2 }^{ { 2 }^{ 1024 } }+1 \right) \\ =\left( 2-1 \right) \left( 2+1 \right) \left( 4+1 \right) \left( 16+1 \right) \left( 256+1 \right) .....\left( { 2 }^{ { 2 }^{ 1023 } }+1 \right) \left( { 2 }^{ { 2 }^{ 1024 } }+1 \right) \\ \\ Multiply\quad (2-1)\quad and\quad (2+1)\\ \\ =\left( 4-1 \right) \left( 4+1 \right) \left( 16+1 \right) \left( 256+1 \right) .....\left( { 2 }^{ { 2 }^{ 1023 } }+1 \right) \left( { 2 }^{ { 2 }^{ 1024 } }+1 \right) \\ \\ Multiply\quad (4-1)\quad and\quad (4+1)\\ \\ =\left( 16-1 \right) \left( 16+1 \right) \left( 256+1 \right) .....\left( { 2 }^{ { 2 }^{ 1023 } }+1 \right) \left( { 2 }^{ { 2 }^{ 1024 } }+1 \right) \\ \\ It\quad will\quad keep\quad on\quad breaking\quad till\quad it\quad reaches\quad this-\\ \\ =\left( { 2 }^{ { 2 }^{ 1024 } }-1 \right) \left( { 2 }^{ { 2 }^{ 1024 } }+1 \right) \\ =\left( { \left( { 2 }^{ { 2 }^{ 1024 } } \right) }^{ 2 }-1 \right) \\ =\left( { 2 }^{ { 2 }^{ 1024 }.2 }-1 \right) \\ =\left( { 2 }^{ { 2 }^{ 1024+1 } }-1 \right) \\ =\left( { 2 }^{ { 2 }^{ { 2 }^{ 10 }+1 } }-1 \right)$