How to prove that a set A is finite iif it is equipotent to $J_n=\{1,…,n\}$ for some $n{\in}\mathbb{N}$? Considering the (not most common) definition:

A set is infinite if it is equipotent to a proper subset of itself. A set is finite if it is not infinite.

How can I prove that a set $A$ is finite iif it is equipotent to $J_n=\{1,…,n\}$ for some $n{\in}\mathbb{N}$ (assuming that I already proved that $J_n$ is a finite set for every $n{\in}\mathbb{N}$)?
 A: Suppose that $A$ is not equipotent to $J_n$ $\forall n$. So you can find a surjective function $f:A\rightarrow\mathbb N$. The idea is that if the function is not surjective than you can find a bijection between $A$ and $J_n$ where $n$ is the maximum of $Im(f)$. So $A$ has at least the cardinality of $\mathbb N$, that has the cardinality of the even numbers that is a proper subset of it, so $A$ is infinite.
A: Suppose $A$ is not equipotent with any of the $J_n$. Then we see that we can produce a surjection from $A$ to $\mathbb{N}$. If not then there is some maximum $n\in \mathbb{N}$ for $f(A)$. From there you can whittle it down to a bijection to a $J_n$. Then $A$ you can conclude by contradiction that $A$ is equipotent to some $J_n$. 
For the reverse direction, if $A$ is equipotent to $J_n$ for some $n$ then there is a bijection $f:A\to J_n$. Pick any $k\in J_n$ then there is no bijection from $J_n$ to $J_n\setminus \{k\}$. Since $f$ is a bijection, argue that there is no bijection from $A$ to $A\setminus\{f^{-1}(k)\}.$
