(Common Lisp) Tabulate k = 2^n Minimizing Multiplication

Question

Given the following problem:

;3.3 Tabulate the function k = 2^n for n = 1..50. 
;Do this for the fewest possible multiplications.[3]

I wrote this answer:

(defun k (n) (ash 2 (1- n)))

(loop for n from 1 to 50 do
      (format t "k(~a) = ~a ~%" n (k n)))

What do you think?

EDIT: based on feedback received, (ash 2 (1- n)) has been simplified to (ash 1 n):

(defun k (n) (ash 1 n))

(loop for n from 1 to 50 do
      (format t "k(~a) = ~a ~%" n (k n)))

Answer 1

If you use 1 instead of 2 as the number to shift, you won't have to subtract one from n. I.e. you can just define k as (ash 1 n).

Other than that your code looks fine.