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

    return fastPow(x, n);/* Time O(log(N)) | Space O(log(N)) */
}
Fast Power - Recursive 
Time O(log(N)) | Space O(log(N))
https://leetcode.com/problems/powx-n/
@param {number} x
@param {number} n
@return {number}