Knight is might!

Relevant wiki: Coloring (Parity Arguments)

In one move we know that a knight moves from a black square to a white one or the other way around, which means that it alternates its colour after each move. So, we can conclude that after every 2 moves, that is, after an even number of moves, the knight ends up on the same colour that it started from. Now, if it must touch all 64 squares just once and end up at H8, then it will require 63 moves, which is an odd number. An odd number of moves will result in the knight ending up on a white square as it started on A1 (black).

But H8 is black.

So the knight can never ever reach H8, by going through each and every square.

Relevant wiki: Coloring (Parity Arguments)

$\text{A knight always moves to its opposite color.}$

Now the knight start from a1.

While visiting every square exactly once, the walk will have exactly 63 moves.

With odd number of moves, knight can only end up on a white colored square (starting from black). Hence it can never reach h8 which is black colored.

Because the knight is already on a square, it will need to move 63 times to cover every square on the board.

Every knight move will put the knight on an opposite color. The knight is on a black square, so after one move the knight will be on a white square, after two the knight will be on a black square, etc. Therefore, after an odd number of moves, the knight must be on a white square.

The final square is black. It is impossible for the knight to be on this square in 63 moves because 63 is odd. Then, the knight cannot be on this square after it covers the whole board.

This will apply to any square board with an even side length. The total number of squares would be $n \times n$ , where $n$ is an even number, so there would be an even number of squares. WLOG, the knight is on the bottom left black square (so the goal is the top right black square). The knight would have to make $n^2-1$ total moves, which is odd, but after an odd number of moves the knight would be on a white square, so it is impossible for the knight to be in the opposite corner.