I guess it's worth saying this in an answer: I don't think this is a meaningful question.  Consider the function defined recursively by $f(1) = 1, f(n+1) = 2 f(n)$.  Clearly $f(n) = 2^n$ for all $n$.  You shouldn't consider this formula an "escape" from the recursive definition, for the very simple reason that the exponential function is usually **defined** by this very recursion!  (One can, of course, do something incredibly silly like define $2^n$ to be $e^{n \ln 2}$.  Whether you think this constitutes an "escape" from the recursive definition is up to you, but what it doesn't constitute is a fast method to compute powers of two.)

What you can ask for, instead, is an algorithm faster than the naive one above.  There is, in fact, such an algorithm; it goes by the name <a href="http://en.wikipedia.org/wiki/Exponentiation_by_squaring">binary exponentiation</a> or exponentiation by squaring, and it basically works by replacing the recursion $f(n+1) = 2f(n)$ by a pair of recursions
$$f(2n) = f(n)^2, f(2n+1) = 2f(n)^2.$$

Does this constitute an "escape" from the recursive definition?  I don't know; it still requires that you compute some smaller powers of two, just not as many.