Well, let's think it through when the pirates don't like bloodshed, borrowing miranche's condensed format:
Two pirates. Pirate #2 can only get #1's vote if he gives all the loot to #1, so #2 ends up with nothing and #1 gets all the gold.
Three pirates. #3 must win one vote. He knows #1 will vote against any proposal because #1 gets everything in a 2-pirate split. However, #2 will vote for him even if #3 keeps all the gold, since #2 won't end up with any gold regardless, and wishes to avoid #3 being killed. Hence, #3 keeps all the gold.
Four pirates. #4 must get two votes. He knows #3 will vote against, as he gets it all in a 3-pirate split. But #4 can keep all the gold because #2 and #1 will vote for him regardless to avoid bloodshed. After all, they won't get any gold anyway if #4 is killed.
Five pirates. #5 needs two votes. He knows #4 will vote against, as he'd get it all in a 4-pirate split. But under these rules #5 can keep all the gold, because in a 4-pirate split number #1, #2, and #3 won't get any gold and will therefore vote in favor to avoid bloodshed.
So you're right! In fact, with these rules it doesn't matter how many pirates there are (above 2) -- the first pirate can keep all the gold. The puzzle is not so interesting this way, is it?