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 $f$ is unbounded above.

Proof
If $n = 1$, then $f$ is the identity function.

By the Archimedean Principle, the real numbers are unbounded above.

Thus by definition of the identity function: $f$ is unbounded above.

Let $n \ge 2$.

Aiming for a contradiction, suppose that $f$ is bounded above by $b \in \R$.

suppose that $b > 0$.

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

Let $x \gt b$.

Then:
 * $\dfrac{x^n - b} {x - b} \le 0$

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:
 * $f' \left({p}\right) = n p^{n - 1}$

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

By Power of Strictly Positive Real Number is Positive and Positive Real Numbers Closed under Multiplication it follows that:
 * $p \gt 0 \implies n p^{n - 1} \gt 0$

But this is impossible because it was previously established that $p \ge b \gt 0$.

From this contradiction it follows that there can be no such $b$.

Hence the result.

Also see

 * Strictly Positive Integer Power Function Strictly Succeeds Each Element