Real Power is of Exponential Order Epsilon

Theorem
Let:


 * $f: \left[{0\,.\,.\,\to}\right) \to \R: t \mapsto t^r$

be $t$ to the power of $r$, for $r \in \R, r > -1$.

Then $f$ is of exponential order $\epsilon$ for any $\epsilon > 0$ arbitrarily small in magnitude.

Proof
For $t > 0$, $t^r$ is continuous.

At $t = 0$, defining $0^r = 0$, the function is continuous from the right:

Fix $t > 1$.

By the Archimedean Principle, there is a natural number $n$ such that $n > r$.

By Real Power is Strictly Increasing, $t^r < t^n$.

Recall from Polynomial is of Exponential Order Epsilon, $t^n < Ke^{at}$ for any $a > 0$, arbitrarily small in magnitude.

Therefore the inequality $\ t^r < Ke^{at}$ has solutions of the same nature.