Let $n \in \mathbb{N}$ and consider the set of nonstandard natural numbers $^*\mathbb{N}$ in sense of nun standard analysis.
I want to show that for each $m \in (^*\mathbb{N}) \backslash \mathbb{N} $ we have $n <^* m$ where $<^*$ is the extension of $<$ to nonstandard natural numbers.
My ideas:
I tried to consider the set $A:= \{a \in \mathbb{N} \vert a < n\}$ and it's extension $^*A$ in order leading it to a contradiction or maybe directly by showing $^*A \subset \mathbb{N}$ using transfer principle.
But till now I don't get how to construct it.
For a fixed natural number $n$, consider the first order formula $$\forall x:(x=0)\lor(x=1)\lor\dots\lor(x=n) \lor (x>n) $$ This is valid in $\Bbb N$, so it also holds in ${}^*\Bbb N$.
In particular, if the given $m\in{}^*\Bbb N$ is not in $\Bbb N$, it follows that $m>n$.