Bare-bones nth root of a prime

nthroot.py

def pow(x, n):
    """
    Bare-bones 'raise x to an integer power n'
    """
    y = 1
    for _ in range(n):
        y *= x
    return y

def nthroot(p, n):
    """
    Minimal implementation of 'nth root of p'

    Uses a fixed-point iteration for x ** -n = 1/p
    """
    # Initial guess
    x = 1.5

    # Iteration
    for _ in range(1000):
        x += 1 / pow(x, n) - 1 / p

    # Return the fixed point solution
    return x

# Test with primes and their nth roots
for p in [2, 3, 5, 7, 11, 13]:
    for n in range(2, 11):
        reference = p ** (1 / n)
        ours = nthroot(p, n)
        print(f"{p} ** (1 / {n}) =", reference, "?=", ours, ": error =", ours - reference)