I'll consider the problem of computing $x^\frac1{q}, \; q > 0$; as I've already mentioned in the comments, one can decompose any positive rational number as $m+\dfrac{p}{q}$, where $m,p$ are nonnegative integers, $q$ is a positive integer, and $p < q$. Thus for computing $x^{m+\frac{p}{q}}$, one could use binary exponentiation on $x^m$ and $\left(x^\frac1{q}\right)^p$ and multiply the results accordingly.
A.N. Khovanskiĭ, in his book on continued fractions, displays a continued fraction representation for the binomial function:
$$(1+z)^\alpha=1+\cfrac{2\alpha z}{2+(1-\alpha)z-\cfrac{(1-\alpha^2)z^2}{3(z+2)-\cfrac{(4-\alpha^2)z^2}{5(z+2)-\cfrac{(9-\alpha^2)z^2}{7(z+2)-\cdots}}}}$$
which converges for $|\arg(z+1)| < \pi$.
Letting $z=x-1$ and $\alpha=\dfrac1{q}$, one can then evaluate this continued fraction (with, say, Lentz-Thompson-Barnett) to generate a "seed" that can be subsequently polished with Newton-Raphson, Halley, or any of a number of iterations with high-order convergence. You'll have to experiment with how accurate a seed you need to start up the iteration, by picking a not-too-small tolerance when evaluating the continued fraction.
Here's some Mathematica code demonstrating what I've been saying earlier, for computing $\sqrt[3]{55}$:
With[{q = 3, t = 55, prec = 30},
y = N[2 + (1 - 1/q) (t - 1), prec];
c = y; d = 0; k = 1;
While[True,
u = (k^2 - q^-2) (t - 1)^2; v = (2 k + 1) (t + 1);
c = v - u/c; d = 1/(v - u d);
h = c*d; y *= h;
If[Abs[h - 1] <= 10^-4, Break[]];
k++];
FixedPoint[
Function[x, x ((1 + q) t - x^q (1 - q))/(x^q (1 + q) - (1 - q) t)],
1 + 2 (t - 1)/q/y]]
Here, I've arbitrarily chosen to stop when the continued fraction has already converged to $\approx 4$ digits, and then polished the result with Halley's method. The result here is good to $\approx 28$ digits. Again, you'll have to experiment on the accuracy versus expense of evaluating the "seed", as well as picking the appropriate iteration method for polishing the seed.
Factorization and arithmetic are two wildly different subjects.
The complexity of arithmetic is reasonably well understood. You might think that arithmetic (say addition, subtraction, multiplication, division, and raising to a power) is trivial. But multiplying large numbers is non-trivial. The algorithm used in practice to multiply really large numbers (Schönhage-Strassen) was only invented in 1971, and recently (2007) an even better algorithm was suggested by Fürer, though his algorithm isn't faster in practice even for huge numbers.
It gets even worse for factorization. There are various "tricks" used to factor numbers, and they use non-obvious algebraic number theory. The two most wildly-used algorithms today are Lenstra's elliptic curve factorization and the number field sieve (the number fields in question are $\mathbb{Q}$ adjoined with the root of some high-degree polynomial).
Wikipedia has a nice summary of best known results. There are also books on the subject, check Wikipedia's page on computational number theory, or (for a different selection of topics), Algebraic Complexity Theory.
Best Answer
$$ \frac{52! - 51!}{51! - 50!} = \frac{51!(52-1)}{50!(51-1)} = \frac{51(52-1)}{51-1} = \cdots $$