classic recursion (watch this, it blows stack)
(defn exp [x n]
     (if (zero? n) 1
         (* x (exp x (dec n)))))
tail recursion
(defn exp [x n]
  (loop [acc 1 n n]
    (if (zero? n) acc
        (recur (* x acc) (dec n)))))
functional
(defn exp [x n]
  (reduce * (repeat n x)))
sneaky (also blows stack, but not so easily)
(defn exp-s [x n]
  (let [square (fn[x] (* x x))]
    (cond (zero? n) 1
          (even? n) (square (exp-s x (/ n 2)))
          :else (* x (exp-s x (dec n))))))
library
(require 'clojure.contrib.math)
Clojure has a power function that works well: I'd recommend using this rather than going via Java interop since it handles all the Clojure arbitrary-precision number types correctly. It is in namespace clojure.math.numeric-tower.
It's called expt for exponentiation rather than power or pow which maybe explains why it's a bit hard to find ... anyway here's a small example (note that use works but better use require):
(require '[clojure.math.numeric-tower :as math :refer [expt]])  ; as of Clojure 1.3
;; (use 'clojure.contrib.math)     ; before Clojure 1.3
(expt 2 200)
=> 1606938044258990275541962092341162602522202993782792835301376
You must first install the Java package org.clojure.math.numeric-tower to make the Clojure namespace clojure.math.numeric-tower accessible!
On the command line:
$ lein new my-example-project
$ cd lein new my-example-project
Then edit project.clj and add [org.clojure/math.numeric-tower "0.0.4"] to the dependencies vector.
Start a lein REPL (not a clojure REPL)
$ lein repl
Now:
(require '[clojure.math.numeric-tower :as math])
(math/expt 4 2)
;=> 16
or
(require '[clojure.math.numeric-tower :as math :refer [expt]])
(expt 4 2)
;=> 16
You can use java's Math.pow or BigInteger.pow methods:
(Math/pow base exponent)
(.pow (bigdec base) exponent)
When this question was originally asked, clojure.contrib.math/expt was the official library function to do this. Since then, it has moved to clojure.math.numeric-tower
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