Using a Lambda

We can solve this by using a lambda expression. mod-expt can be defined as:
     (define mod-expt
       (lambda (base exponent modulus)
         (together-copies-of
           (lambda (x y)
             (remainder (* x y) modulus))
           exponent base)))
Note the similarity to:
     (define power
       (lambda (base exponent)
         (together-copies-of 
           *
           exponent base)))