Let $A$ be a set inside $\Bbb N^*$ that contains $1$ such that : $$i) ∀n ∈ A, 2n ∈ A$$ and $$ii) ∀n ∈ \Bbb N^* , n + 1 ∈ A ⇒ n ∈ A.$$
I was asked to show that $∀m ∈\Bbb N , 2^m ∈ A$ which was straightforward with induction but then I had to show that : $$A=\Bbb N^*$$ I tried using induction again, to show that $\Bbb N^*$ is included in $A$ thus the result, since this exercise is in the induction chapter but didn't get anywhere.
$1\in A$ and the property (i) gives $\{2^n:n=0,1,2,3,\dots\}\subseteq A$.
Now the property (ii) gives that, $n\in A$ $\implies m\in A,\forall m\in\Bbb{N}^*$ such that $m\leq n$.
Now let $n\in \Bbb{N}^*$ then $n< 2^n\in A$. Which gives $n\in A$. Thus $\Bbb{N}^*\subseteq A$.