<p>What is the time complexity? Why?</p>

<pre><code>(define (mult a b)
      (define (internal a accum)
        (if (= a 1) accum
            (internal (- a 1) (+ accum b))))
      (internal a b))

(define (to-the-power-of m n)
  (define (internal x accum)
    (if (= x 0) accum
        (internal (- x 1) (mult accum m))))
  (internal n 1))
</code></pre>

Assuming addition and multiplication are both counted as a single operation, this function performs O(m^n) operations.
First consider the mult function. It (mult a b) will perform exactly a-1 additions. Since, the asymptotic growth is the same, lets approximate this by a, for mathematical simplicity.
Now for the to-the-power-of function, this performs n calls to the mult function. These calls are to (mult 1 m), yield m, then to (mult m m), yielding m^2, then to (mult m^2 m), yielding m^3 and so on upto m^n. So the total number of operations performed here is the sum m^0 + m^1 + ... + m^n. This is (m^n - 1) / (m-1) which grows as m^n.