Can I prove $2^{-n}$ converges to $0$ without invoking the Axiom of Completeness or the Archimedean property of the reals? Is it possible to show that for any real number $x > 0$, there exists a natural number $n$ such that $2^{-n} < x$ without invoking the Axiom of Completeness or the Archimedean property of the reals?
 A: You haven't specified what properties you are allowing to be used.
It's pretty easy to find ordered fields where such a sequence can't converge to zero. For example, take any ordering on the field of rational functions $\mathbb{R}(x)$ i.e. the field of all functions that can be written as the quotient of two polynomials. Without loss of generality, we can take $x$ to be a positive and infinite; that is, $x > a$ for all real numbers $a$.
In this field, we always have $0 < 1/x < 2^{-n}$ for every natural number $n$, so the sequence clearly cannot converge to zero.
There are even worse examples; in some ordered fields, the only convergent sequences are the ones that are eventually constant! (where "sequence" is limited only to sequences indexed by natural numbers)
(if you replace the field of rational functions with some other suitable example, then all of the above applies to real closed fields as well, which is basically just a completeness axiom suitable for working with polynomials)
A: If for all $\varepsilon>0$ there exists $N_{\varepsilon}\in\Bbb N$ such that $2^{-N_{\varepsilon}}<\varepsilon$, then for all $\alpha>0$ you have $2^{N_{1/\alpha}}>M$. Which proves the archimedean property. So I guess you cannot avoid it. Completeness can be avoided and, in fact, if you try to prove it as usual you won't ever need to use the fact that bounded sets have a least upper bound.
A: It depends on how you define convergence.
If you define $(x_n)_{n\in\mathbb{N}}$ to converge to $a$ if, for every positive rational $\varepsilon$, there exists $k$ such that, for $n>k$, $|x_n-a|<\varepsilon$, then you don't need the Archimedean property.
Indeed, it's easy to show that, for every natural $n$, $2^n>n$ and the argument can be quickly completed.
However, you need the Archimedean property for proving that, given a positive real $\varepsilon$ there exist a rational $q$ such that $0<q<\varepsilon$. This is necessary for the above definition to be equivalent to the one with “for every positive real $\varepsilon$”.
A: The "Archimedean property" is already present in the natural numbers, and can be proven from the Peano axioms. It therefore makes no sense to avoid making use of it. 
Let an $x\in{\mathbb R}_{>0}$ be given, and put $$n:=\left\lceil{1\over x}\right\rceil\geq{1\over x}\ .$$ The set $\{1,2,3,\ldots,2^n\}$ contains the $n+1$ numbers $1$, $2$, $4$, $8$, $\ldots$, $2^n$. It follows that $2^n>n$, hence
$$0<2^{-n}<{1\over n}\leq x\ .$$
A: For $\epsilon\in\mathbb{R}$ be a real number such that $\epsilon > 0$ 
let $N_{\epsilon}\in\mathbb{N}$ be the smallest integer such that
$N_{\epsilon} > ln(\frac{1}{\epsilon})/ln(2)$.
Then for all $n>N_{\epsilon}$, $2^{-n}<\epsilon$.
