User blog comment:Ubersketch/My unnamed function is finished?/@comment-5529393-20180921025306/@comment-30754445-20180921144700

Looks like you're right.

What's interesting, though, is that Deedlit's statement remains true for other choices of f as well. You can set f to be 10n or 2^n or n! or even n[n arrows]n, and it won't affect the end result by much.

(basically you're adding the ordinal that corresponds to your choice of f to w^w,  and due to a peculiarity of ordinal addition, a+w^w = w^w for all a<w^w).