Strictly Positive Integer Power Function is Unbounded Above

Theorem
Let $\R$ be the real numbers with the usual ordering.

Let $n \in \mathbb N_{>0}$.

Let $f\colon \R \to \R$ be defined by $f\left({x}\right) = x^n$.

Then the image of $f$ is unbounded above in $\R$

Proof
If $n=1$, then $f$ is the identity function. Since the reals are unbounded above, $f\left({\R}\right)$ is unbounded above.

Suppose now that $n \ge 2$.

Suppose that $f\left({\R}\right)$ is bounded above by $b \in \R$.

Assume without loss of generality that $b > 0$.

Then by the definition of an upper bound, for any $x$, $x^n \le b$.

Thus for $x > b$, $\displaystyle \frac{x^n - b}{x-b} \le 0$.

By the Mean Value Theorem, there must be a point $p$ between $b$ and $x$ such that
 * $\displaystyle f'\left({p}\right) = \frac{x^n - b}{x-b}$.

By Derivative of Power,
 * $f'\left({p}\right) = n p^{n-1}$

Thus $n p^{n-1} \le 0$.

But since $p \ge b > 0$, this is impossible.