Strictly Increasing Sequences of Natural Numbers Dominate Their Index

Let $\mathbb{N}$ denote the natural numbers with the addition of that definition and the order $\le$, and let $(n_k)_{k\in\mathbb{N}}$ be a sequence in $\mathbb{N}$ that is strictly increasing in the sense of Subsequence of a Sequence in a Set.

Then $k\le n_k$ for every $k\in\mathbb{N}$. Consequently, for every $N\in\mathbb{N}$ there exists $k\in\mathbb{N}$ with $N\le n_k$.