No, f = O(n log n) means that f doesn't grow significantly faster than n log n. That is true for f = n log n, but it's also true for f = n. The "or better" is implicit by using big-O.

Note that this wouldn't be true if the standard said that it has to be Θ(n log n).

Big O complexity is an upper bound.

If something is O(1) it's also O(n) and O(n!), since it's an upper bound.