Hint, not a solution
@Placholdr I'm not sure if this will work, but here's an approach:
Use contraction. Assume a_n *is* a perfect square. Consider all possibilities for the square root's last digit. It can't be even, because squaring an even number gives another even number. That leaves 1, 3, 5, and 9. ***1^2 (that is, squaring a number that ends in 1) ends in 1. ***3^2 ends in 9, ***5^2 ends in 5, and ***9^2 ends in 1. So the proposed square root must end in 1 or 9. Explore those possibilities and see if you can eliminate them.
Alternately, maybe the old algorithm for taking square roots by hand is useful.
Again, these may be dead ends.
Hint, not a solution
I agree, this is probably the best method.
Another observation I made was that perfect squares can be expressed as the sum of odd numbers, so maybe that can help since the sequence /({a_k}/) can be described as /(\sum\limits_{i=0}^{2k-1}10^{i}/).
Step(s) towards solution
Alright, I figured it out based on @peterdrake 's hint. You'll find that any integer ending either in 1 or 9 squared has to be a number divisible by 20 plus 1 (20a+1 where a is a natural number). This can be done by multiplying out \((10n+1)^2\) and \((10n+9)^2\) where \(n\) is a natural number. It should be pretty straight forward from there.
Since no number from the sequence \(\{a_k\}\) follows this, then by contradiction, \(\{a_k\}\) cannot be a perfect square.
Hint, not a solution
@sojournTime @peterdrake @Placholdr
Another hint, or rather a route to explore: modular arithmetic is your friend. Both for square numbers and your string of 1s, look for the possible remainders modulo several different numbers. (Then try to prove the pattern you spot, of course...)