(x^x)^(x^x), x^(x^(x^x)), etc...

[Russ Cox]

I think the correspondence works like this.

The single node tree is "x".  Making a node f2
a child of f1 represents f1^f2.  Since (f1^f2)^f3
is just f1^(f2*f3) we can think of it as f1 raised
to both f2 and f3, that is, f1 with f2 and f3 as 
children.

So the functions you posted are (sideways)

o--o--o--o (f1)

o--o--o (f2)
 \
  o

o--o--o (f3)
    \
     o

  o
 /
o--o (f4)
 \
  o

o--o (f5)
 \
  o--o

Russ