Natural Number Power is of Exponential Order Epsilon

Theorem
Let $n \in \N$ be a natural number.

Then:


 * $t \mapsto t^n$

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

Proof
The proof proceeds by induction on $n$, where $n$ is the degree of the polynomial.

Basis for the Induction
When $n = 0$, the mapping is a constant mapping.

By Constant Function is of Exponential Order Zero and Raising Exponential Order, the mapping is of exponential order $\epsilon$.

This is the basis for the induction.

Induction Hypothesis
Fix $k \in \N$ with $k \ge 0$.

Assume:


 * $t^k \in \mathcal{E}_\epsilon$

That is,


 * $\forall t \ge M : \left \vert {t^k}\right \vert < Ke^{\epsilon t}$

For some $M > 0$, $K > 0$, and for any $\epsilon > 0$ arbitrarily small.

This is our induction hypothesis.

Induction Step
We have:


 * $x^k$ is of exponential order epsilon by the induction hypothesis


 * Identity is of Exponential Order Epsilon


 * Product of Functions of Exponential Order

Therefore, $x^{k + 1}$ is also of exponential order epsilon.

The result follows by the Principle of Mathematical Induction.