Super Pow - Implementation

Here's the recursive breakdown in code:

MOD = 1337

def power(a, n, mod):
    result = 1
    a = a % mod
    while n > 0:
        if n % 2 == 1:
            result = (result * a) % mod
        n = n // 2
        a = (a * a) % mod
    return result

def superPow(a, b):
    if not b:
        return 1
    
    last_digit = b[-1]
    remaining = b[:-1]
    
    # a^b = (a^remaining)^10 * a^last_digit
    part1 = power(superPow(a, remaining), 10, MOD)
    part2 = power(a, last_digit, MOD)
    
    return (part1 * part2) % MOD

Each recursive call shrinks $b$ by one digit. The base case is when $b$ is empty, which returns 1.