There will be an answer.

Japanese Programmer's: From recursion to iteration.

I found an answer.

(define (fast-expt2 b n)
  (fast-expt-iter 1 b n))

(define (fast-expt-iter a b n)
  (cond ((= n 0) a)
        ((even? n) (fast-expt-iter a (square b) (/ n 2)))
        (else (fast-expt-iter (* a b) b (- n 1)))))