Strictly Increasing Sequence of Natural Numbers

Theorem
Let $\N^*$ be the set of natural numbers without zero: $\N^* = \left\{{1, 2, 3, \ldots}\right\}$.

Let $\left \langle {n_r} \right \rangle$ be a sequence in $\N^*$.

Let $\left \langle {n_r} \right \rangle$ be strictly increasing.

Then $\forall r \in \N^*: n_r \ge r$.

Proof
This is to be proved by induction on $r$.

For all $r \in \N^*$, let $P \left({r}\right)$ be the proposition $n_r \ge r$.

Basis for the Induction
When $r = 1$, it follows that $n_1 \ge 1$ as $\N^*$ is bounded below by $1$.

Thus $P(1)$ is true.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So our induction hypothesis is that $n_k \ge k$.

Then we need to show that $n_{k+1} \ge k+1$.

Induction Step
This is our induction step:

Suppose that $n_k \ge k$.

Then as $\left \langle {n_r} \right \rangle$ is strictly increasing:
 * $n_{k+1} > n_k \ge k$

From Precedes Next (which applies because the Natural Numbers are a Naturally Ordered Semigroup), we have:
 * $n_{k+1} > k \implies n_{k+1} \ge k+1$

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore $\forall r \in \N^*: n_r \ge r$.