Root in root in root in root in root in root.

This looks messy to begin with, but, surprisingly, $\sqrt{1+793 \times 795} = 793$ ! This is because of the following identity:

$\sqrt{1+(n-1)(n+1)} = \sqrt{1+n^2-1} = n$ .

This works out perfectly, because applying this recursively allows us to 'climb' the radicals by incrementing our products! So,

$\sqrt{1+788\sqrt{1+789\sqrt{1+790\sqrt{1+791\sqrt{1+792\sqrt{1+793 \times 795}}}}}}$

$=\sqrt{1+788\sqrt{1+789\sqrt{1+790\sqrt{1+791\sqrt{1+792 \times 794}}}}}$

$=\sqrt{1+788\sqrt{1+789\sqrt{1+790\sqrt{1+791 \times 793}}}}$

$=\sqrt{1+788\sqrt{1+789\sqrt{1+790 \times 792}}}$

$=\sqrt{1+788\sqrt{1+789 \times 791}}$

$=\sqrt{1+788 \times 790}$

$= \boxed{789}$ .

1 + 793 x 795 = 1 + 793 ^ 2 + 793 x 2 = (1 + 793)^2 1+ 792 x 794 Note that 794 - 792 = 2. So on, we have 1 + 791 x 793 ...etc. So, 1 + 788 x 790 = (1+788)^2 <=> Result is 789

789

793x795 = (794-1)(794+1)=794^2 - 1 hence 1 + 793x795 = 1 + 794^2 - 1 = 794^2 root of( 794^2) = 794

now

792x794 = (793-1)(793+1)=793^2 - 1 hence 1 + 792x794 = 1 + 793^2 - 1 = 793^2 root of( 793^2) = 793

again

791x793 = (792-1)(792+1)=792^2 - 1 hence 1 + 791x793 = 1 + 792^2 - 1 = 792^2 root of( 792^2) = 792

again

790x792 = (791-1)(791+1)=791^2 - 1 hence 1 + 790x792 = 1 + 791^2 - 1 = 791^2 root of( 791^2) = 791

again

789x791 = (790-1)(790+1)=790^2 - 1 hence 1 + 789x791 = 1 + 790^2 - 1 = 790^2 root of( 790^2) = 790

again

788x790 = (789-1)(789+1)=789^2 - 1 hence 1 + 788x790 = 1 + 789^2 - 1 = 789^2 root of( 789^2) = 789 ANSWER !!!

793 795+1=630436/794=794 792+1=628849/793=793 791+1=627264/792=792 790+1=625681/791=791 789+1=624100/790=790 788+1=622521/789=789

793 795=(794-1) (794+1) 793*795+1=794^2-1+1=794^2 Proceeding like this answer can be obtained.

789