For any such arrangement, we can find nonnegative integers $x_1, x_2, x_3,..., x_{2008}$ such that the obtained number is equal to $1^1\cdot10^{x_1} + 2^2\cdot10^{x_2} +...+2008^{2008}\cdot10^{x_{2008}}$. We know that $1^1\cdot10^{x_1} + 2^2\cdot10^{x_2} +...+2008^{2008} \equiv 1^1 + 2^2 +...+ 2008^{2008}$ (mod 3). We also know that if $a$ is a positive integer, then If $a \equiv 0$ (mod 3), then $a^a \equiv 0$ (mod 3) If $a \equiv 1$ (mod 3), then $a^a \equiv 1$ (mod 3) If $a \equiv 2$ (mod 3), and a is even, $a^a \equiv 1$ (mod 3) If $a \equiv 2$ (mod 3), and a is odd, $a^a \equiv 2$ (mod 3) It follows that $1^1\cdot10^{x_1} + 2^2\cdot10^{x_2} +...+2008^{2008}\cdot10^{x_{2008}} \equiv 2$ (mod 3). But since a perfect square cannot be congruent to 2 (mod 3), it is impossible for any arrangement of the 2008 integers to be a perfect square.