Assumption: If sqrt(2^n- 1) ∈ℕ , it means that there is a k ∈ ℕ such that (2^n-1) = k^2 ⇒ k is odd, so there is m ∈ ℕ such that (2^n-1)=(2m+1)^2 . By solving the equation we get 2^n= 4m^2+4m+2. We divide both sides by 2 for easier observation and then we get 2^(n-1) =2m^2+2m+1. We can see that 2^(n-1) is even for any n>2 and 2m^2+2m+1 is odd for any m∈ℕ. Thus, there is no solution to this equation for n>2, so our assumption is wrong => (2^n)-1 is not a perfect square.