X to the x is not of Exponential Order

Theorem
Let $f: \R \to \R$ be defined on $\left [{0 \,.\,.\, \to} \right)$ with $f \left({x}\right) = x^x$.

Then:
 * $f$ is not of exponential order.

That is, it grows faster than any exponential.

Proof
By the definition of power:
 * $f \left({t}\right) = \exp \left({t \ln t}\right)$

The theorem is equivalent to that there do not exist strictly positive real constants $M$, $K$, $a$ such that:


 * $\forall t \ge M: \left\vert {f \left({t}\right)} \right \vert < K e^{a t}$

Aiming for a contradiction, assume such constants $M$, $K$ and $a$ exist.

From the lemma, there exists a constant $C$ such that:
 * $\forall t > C: \left\vert {f \left({t}\right)} \right \vert > K e^{a t}$

For any $M$ chosen, $M + C > C$, thus from the lemma:
 * $\left\vert {f \left({M + C}\right)}\right\vert > K e^{a t}$

However, $M + C > M$, thus from the assumption:
 * $\left\vert {f \left({M + C}\right)}\right\vert < K e^{a t}$

Therefore, a contradiction has arisen.