Upper Bounds for Prime Numbers/Result 1

Theorem
Let $p: \N \to \N$ be the prime enumeration function.

Then $\forall n \in \N$, the value of $p \left({n}\right)$ is bounded above.

In particular:
 * $\forall n \in \N: p \left({n}\right) \le 2^{2^{n-1}}$

Proof
Proof by strong induction:

Let us write $p_n = p \left({n}\right)$.

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:
 * $p \left({n}\right) \le 2^{2^{n-1}}$

Basis for the Induction
$P(1)$ is true, as this just says $2 \le 2^{2^0} = 2$.

This is our basis for the induction.

Induction Hypothesis
Suppose that, for some $k \in \N$, each of $P \left({1}\right), P \left({2}\right), \ldots, P \left({k}\right)$ is true.

It remains to show that it logically follows that $P \left({k+1}\right)$ is true.

That is, that:
 * $p \left({k+1}\right) \le 2^{2^k}$

This is our induction hypothesis.

Induction Step
This is our induction step:

The result follows by the Second Principle of Mathematical Induction.

Note
It can be seen that the limit found is wildly extravagantly large.

However, it is an easily established result, and it has its uses.