Proof of Strictly Increasing Sequences of Natural Numbers Dominate Their Index

All properties of the order on $\mathbb{N}$ used below are those listed in Properties of the Order on the Natural Numbers, and $m+1=S(m)$ for the successor map $S$ is claim 1 of Arithmetic of Addition on the Natural Numbers.

For $k\in\mathbb{N}$ let $P(k)$ be the assertion $k\le n_k$. We prove $P(k)$ for every $k$ by the principle of induction.

Base case. $1\le n_1$ by claim 4 of Properties of the Order on the Natural Numbers.

Inductive step. Assume $k\le n_k$. By claim 6 of Properties of the Order on the Natural Numbers, $S(k)\le S(n_k)$, that is $k+1\le n_k+1$.

By the strict increase hypothesis of Subsequence of a Sequence in a Set we have $n_k<n_{k+1}$, so by claim 7 of Properties of the Order on the Natural Numbers there is $t\in\mathbb{N}$ with $n_{k+1}=n_k+t$. By claim 4, $1\le t$, and hence by claim 6, $n_k+1\le n_k+t=n_{k+1}$.

Combining $k+1\le n_k+1$ and $n_k+1\le n_{k+1}$ with transitivity (claim 1) gives $k+1\le n_{k+1}$, which is $P(k+1)$.

By induction, $k\le n_k$ for every $k\in\mathbb{N}$.

For the final assertion, let $N\in\mathbb{N}$ and take $k=N$; then $N\le n_N$.