#lang racket

; Let's unwrap the generic formulation of above and write it directly
(define fn-cc (lambda (x)

  (define fn-cc-r (lambda (x cc)
     (if (<= x 1) 
         ; base case is 1
         (cc 1)
         ; recursion is x * (fn(x-1))
         (fn-cc-r (- x 1) (lambda (result) (cc (* x result))))
         )
  ))

  (fn-cc-r x (lambda (result) result))
)

(fn-cc 3)