Upper Bounds for Prime Numbers/Result 2

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

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

Let us take $N = p_1 p_2 \cdots p_n + 1$.

By the same argument as in Euclid's Theorem, we have that either $N$ is prime, or it is not.

If $N$ is prime, then either $N = p_{n+1}$ or not, in which case $N > p_{n+1}$.

In the second case, $N$ has a prime factor not in $\left\{{p_1, p_2, \ldots, p_n}\right\}$

Therefore it must have a prime factor greater than any of $\left\{{p_1, p_2, \ldots, p_n}\right\}$.

In any case, the next prime after $p_n$ can be no greater than $p_1 p_2 \cdots p_n + 1$.

Hence the result.

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.