Let n = 3, so there are 3 gentlemen:
1, 2, 3
Then all possible (equally likely) permutations of hats are:
1, 2, 3 (1,2,3 get their hats)
1, 3, 2 (1 gets his hat)
2, 1, 3 (3 gets his hat)
2, 3, 1 (noone gets his hat)
3, 1, 2 (noone gets his hat)
3, 2, 1 (2 gets his hat)
Probability of 1 gentleman getting his hat:
4/6 (from example above)
4/6 ≠ 1/n! = 1/3! = 1/6
Probability n gentleman getting their hats, i.e. everyone gets their hat (if I understand you correctly):
1/6 (from example above)
1/6 ≠ 1/(n-1)! = 1/(3-1)! = 1/2! = 1/2
But,
1/6 = 1/n! (the only permutation where all elements are in their positions)
No?