Strictly Positive Integer Power Function is Unbounded Above

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

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

Let $f: \R \to \R$ 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 real numbers are unbounded above, $f \left({\R}\right)$ is unbounded above.

Let $n \ge 2$.

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

Assume WLOG that $b > 0$.

Then by the definition of an upper bound:
 * $\forall x \in \R: x^n \le b$

Thus for $x > b$:
 * $\dfrac{x^n - b}{x-b} \le 0$

By the Mean Value Theorem, there exists a point $p$ between $b$ and $x$ such that:
 * $f' \left({p}\right) = \dfrac{x^n - b}{x-b}$

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.

Also see

 * Strictly Positive Integer Power Function Strictly Succeeds Each Element