Consider 2 cases:
1. Dn = 6
2. Dn = 0
In case 1 if the last digit of N is 6 i.e. Dn=6 then dividing it by 2 gives us a number (N/2) with last digit 3 and hence cannot be a perfect square since any further division of (N/2) will not yield any 2 from it i.e. N consists of a single 2 and hence it is a non-square number.
In case 2 if the last digit of N is 0 i.e. Dn=0 then let the least consecutive sub-sequence of 0s from Dn backwards be upto Di where 1 <= i <= n,since D0 must be 6 to be a (n+1) digited number.
Therefore,from D0 to Di-1 the number consists of mixes of 0s and 6s and from Di to Dn the number consists of all 0s. Pictorially the view is as follows:
Now separate D0 to Di-1 as a string of 0s and 6s and Di to Dn as a string of only 0s.
In other words the number is as follows: N = D0...Di-1 X 10(n-i+1) = 6...6 X 10(n-i+1)
Now by previous argument we know that D0...Di-1 can't be a perfect square and it also consists of a factor 6=2x3. So if the other part i.e "power of 10 is a perfect square[i.e if (n-i+1) is even]" then overall the number N can't be a perfect square.
If (n-i+1) is odd then 10(n-i+1) can be factored as 2n-i+1 X 5n-i+1 where we see that the part 5n-i+1 still can't be a perfect square( as n-i+1 is odd and no other factors of 5 are there in the whole number ).
So in that case also this required result is proved.