Why a full rook tour to h8 is impossible

Can a rook step one square at a time and visit every square exactly once, starting at a1 and finishing at h8? Colour the board. Each single step moves to an adjacent square, which is always the opposite colour.

A path through all $64$ squares makes $63$ steps, so it flips colour $63$ times, an odd number, and must end on the opposite colour to its start. But a1 and h8 are the same colour. So no such path can begin at a1 and end at h8. This is a parity proof: a colouring turns a hopeless search into a one-line argument.