The question you opened
Games in a 128 player bracket
A knockout bracket starts with 128 players, and one loss sends a player home.
How many games does it take to crown a winner?
127 games
Every game sends one player home. For one to remain, 127 have to go, so 127 games.
The bracket needs 127 games. The slow route adds the rounds: 64 games in the first, then 32, 16, 8, 4, 2, and a final, which totals 127. The fast route counts losers. Each game eliminates exactly one player, exactly one player is left standing at the end, so 127 players must be eliminated and 127 games must be played. That reframing is the whole trick, and it generalises immediately.
A single elimination tournament with n players always takes n minus 1 games, whatever n is and whether or not it is a power of two. With 100 players there are byes and uneven rounds, and the answer is still 99. Nothing about the structure matters, because the counting argument never mentions the structure.
This is called a bijection argument: pair each game with the player it eliminates, show the pairing is one to one, then count the easier of the two sets. It is the same move behind the handshake formula, where n people shaking hands with everyone gives n times n minus 1 over 2 handshakes, since each handshake is counted from two sides. Double elimination changes the answer in a predictable way.
Every player except the winner has to lose twice, so the count is roughly 2n minus 2, plus one extra game if the winner of the losers bracket beats the undefeated finalist and forces a rematch. With 128 players that is 254 or 255 games. The habit worth keeping is to ask what each step of a process consumes. When every step removes exactly one unit of something, the number of steps is the number of units, and a page of addition collapses into a subtraction.
Technique: Count what each game removes