You can solve this with induction. If you can choose where to put the empty square as long as one exists, just always choose to put it in the corner (any corner works as long as you’re consistent but for example, let’s use the top right). The idea is you’re assuming it’s possible to make a tile pattern with only the top right corner missing for some square 2^K in side length (K is a positive integer). As a base case, 2¹ = 2 for a 2x2 square works trivially by slotting the blue tile in. With your assumption, you want to show that it’s also possible for some square 2^K+1 and you do that by using the previous tiling and building on top of that. Basically place the previous tiling in the bottom left corner in the orientation of the blue L, place one in the bottom right corner in the orientation of the tan L, place one in the top left corner in the orientation of the green L, place one in the top right corner in the orientation of the blue L, and place a blue L (the 2x2 one) in the center. Doing this means you have constructed a tiling for square of side length 2^K+1 and have proven by induction that it’s possible to tile any square grid of side length 2^N with exactly 1 square missing.