Show that if $2^k\in S\space\space\forall k\in\mathbb{N}$, and if $k\in S$ and $k\ge2$, then $k-1\in S$, then $S=\mathbb{N}$
My attempt:
Let $n\in\mathbb{N},$ hence $2^n\in\mathbb{N}$ by the first assumption. Now denote the set $A$ by the following: $$A:=\{a\in\mathbb{N}: a<2^n\space\text{and}\space a\not \in S\}$$
Because $A\subset\mathbb{N},$ it follows by the Well-Ordering Principle that if $A$ is bounded above it has a maximal element. Call that element $k$. By construction $k+1\notin A$...
Here is where I am stuck. I would like to argue that $k+1\in S$ since $k+1\notin A$ ...
Any help from here would be appreciated!
For every $n \in \mathbb{N}$, if there is a $k \in \mathbb{N}$ such that $2^k = n$, then we know it's in $S$. Otherwise, there exists a $k \in \mathbb{N}$ such that $2^k > n$ as the set of values of $2^k$ has no upper bound. Choose the smallest such $k$ (although you can choose any larger $k$) and call $2^k = m$, where $m$ is in $S$. As $k \in \mathbb{N}$, we have that $m \ge 2$, so we know that $m - 1$ is also in $S$. You repeat this step $m - n$ times, with the $m$ value of each next step being replaced by the $m - 1$ of the previous step, at which time you will get down to $n$, with this being assured to occur as $n \ge 1$ and the steps stop at $m = 2$ giving that $m - 1 = 1$ is in $S$. This shows that all values $n \in \mathbb{N}$ have that $n \in S$, thus giving that $S = \mathbb{N}$.
There is a likely a simpler & more succinct way to show this, but this is not my area of expertise. Also, I hope I'm not misinterpreting anything, nor assuming or using anything I shouldn't be.