Let $S\subseteq \mathbb{N}$.
If

1. $1\in S$
2. for all $k\in \mathbb{N}$, $k\in S⟹(k+1)\in S$

then $S=\mathbb{N}$.

Proof

Suppose for the sake of contradiction that $\mathbb{N}\setminus S\ne \varnothing$, i.e. $\mathbb{N}\ne S$.
By the well-ordering principle, $\mathbb{N}\setminus S$ has a least element $m$.
Since $1\in S$, $m>1$.
Since $m$ a least element of $\mathbb{N}\setminus S$, $(m-1)\in S$
However, by 2), $(m-1)+1\in S$, i.e. $m\in S$, a contradiction.

So $\mathbb{N}=S$.

QED