myPow = (x, n) => {
    if (n < 0) {
        x = (1 / x);
        n = (-n);
    }

    return getPow(x, n);/* Time O(N) */
}
Brute Force - Multiply
Time O(N) | Space O(1)
https://leetcode.com/problems/powx-n/
@param {number} x
@param {number} n
@return {number}