What properties of the positive integers and $1$ guarantee that repeatedly decreasing by $1$ eventually leads to $1$? 
What properties of the positive integers and $1$ guarantee that decreasing by $1$ eventually leads to $1$?

$f(x)=x-1$
Show that for any positive integer $m>1$, there exists $n$ such that $f^n(m)=1$
I have seen it simply stated that $f(x)<x\quad\forall x\implies\exists\quad  n\mid f^n(x)=1$
But that's not formally sufficient, is it?  We need a few other things.  For example, this argument doesn't state its reliance upon the properties of $\mathbb{N}$ or $1$, but if we were in $\mathbb {Q}$ we might reduce the number $3$ repeatedly by $\frac{2}{3}$ and we would never arrive at $0$.
Firstly, we need to state that $1$ is in some sense the minimum resolution of the integers under addition, in order to show that no step of reducing by $1$ can leapfrog any integer.
Secondly (or perhaps this encapsulates the first) we need to know that the integers are ordered by the function $x\to x+1$ and that $1$ is the least of them by that ordering.
Thirdly we need to state that there are countably many whole numbers, to ensure that we will eventually arrive at $1$.
What is the conventional way of stating these properties, particularly the first?
 A: Firstly, as already pointed out by Ned, you are asking the wrong question; I suggest you start on an introductory logic textbook and learn the meaning of quantifiers and note the difference between:
$
\def\nn{\mathbb{N}}
$

(A) $\exists m \in \nn_{>1}\ \exists n \in \nn\ ( f^n(m) = 1 )$.
(B) $\forall m \in \nn_{>1}\ \exists n \in \nn\ ( f^n(m) = 1 )$.

(A) is what you have in your third line, which is trivial because $2 \in \nn_{>1}$ and $0 \in \nn$ and $f^0(2) = 1$. (B) is what you are actually asking.
Secondly, it is a trivial application of induction, but as your first time encountering formal inductive proofs, it would be instructive for you to write out the argument in full. Here is the outline for you to fill in:

Define $P(m) \equiv ( m>0 \to \exists n \in \nn\ ( f^n(m) = 1 ) )$.
Then $P(0)$ because $\cdots$.
Given any $m \in \nn$ such that $P(m)$:
  $m = 0$ or $m > 0$.   [If you don't have this fact, prove it by induction!]
  If $m = 0$:
    $P(m+1)$ because $f^0(1) = \cdots$.
  If $m > 0$:
    $\exists n \in \nn\ ( f^n(m) = 1 )$.
    Let $c \in \nn$ such that $f^c(m) = 1$.
    Then $c+1 \in \nn$ and $f^{c+1}(m+1) = \cdots$.
    Thus $\exists n \in \nn\ ( f^n(m+1) = 1 )$.
  Therefore $P(m+1)$.
Therefore by induction $\forall m \in \nn\ ( P(m) )$.

And being countable is irrelevant, by the way. The concept generalizes to well-orderings, which are totally-ordered collections such that every element can be decreased only finitely many times (but in a general well-order there is no notion of subtraction). Depending on the specific foundational system, there may be uncountable well-orderings. For the general construction of arbitrarily long well-orderings in ZFC set theory, see this post.
