MHB Evaluate a given expression without using a calculator

[anemone]

Evaluate the expression $(2^{2^{2014}}+1)(2^{2^{2013}}+1)\cdots(2^{2^2}+1)(2^{2^1}+1)(2^{2^0}+1)+1$

 

[kaliprasad]

Evaluate the expression $(2^{2^{2014}}+1)(2^{2^{2013}}+1)\cdots(2^{2^2}+1)(2^{2^1}+1)(2^{2^0}+1)+1$

wish you happy new year

Spoiler

multiply 1st term by $(2^{2^0}-1) $ which is nothing but 1

and using

$(2^{2^{n-1}}+1)(2^{2^{n-1}}-1) = (2^{2^{n}}-1)$ from n = 1 to 2015 one by one
to get

$(2^{2^{2015}}-1+1)$ or $2^{2^{2015}}$

 

[anemone]

wish you happy new year

Spoiler

multiply 1st term by $(2^{2^0}-1) $ which is nothing but 1

and using

$(2^{2^{n-1}}+1)(2^{2^{n-1}}-1) = (2^{2^{n}}-1)$ from n = 1 to 2015 one by one
to get

$(2^{2^{2015}}-1+1)$ or $2^{2^{2015}}$

Thank you kaliprasad for your quick and nice solution! Well done!

Thanks to you too for the New Year wishes...I too wish you and all a safe New Year filled with peace, joy, robust health, happiness, prosperity and all your dreams come true!;)