I'm fairly new to discrete math, and I wasn't sure how to prove the following.
Prove that if $n\geq 2$, then every $2^n \times 2^n$ chessboard can be tiled with non overlapping T-tiles.
If I draw it out, I can clearly see the answer, however I'm not sure how to prove it with induction.
Well, we can divide $2^{n+1}\times 2^{n+1}$ in to four $2^{n}\times 2^{n}$ tables and each can be tiled by induction hypothetis, so you are done.
Clearly base case of induction table $4\times 4$ should not be a problem to see.