How would I prove $10^x> 1$ for $x>0$ without using logarithms? I was asked this question and got completely stumped. The person who asked this has instructed me specifically not to use logarithms. I know there is a more sophisticated way to explain exponentiation using limits of sequences (which would work here), but unfortunately limits are outside the reach in this particular context. So I was wondering, is there a simple way to explain why $0<x<1$ can only be expressed as negative powers of $10$ (or any other number greater than $1$, for that matter) using only high-school algebra? I suppose the proof doesn't have to be entirely rigorous, as long as it's sufficiently convincing.

If $x$ is a positive rational number $p/q$ (where $p$ and $q$ have no common factors), then we can define $10^x$ as
$$
\left(\sqrt[q]{10}\right)^p=\underbrace{\sqrt[q]{10} \cdot \sqrt[q]{10} \cdot \sqrt[q]{10} \cdot \ldots \cdot \sqrt[q]{10}}_{\text{$p$ times}} \, .
$$
For any positive integer $q$, the $q$-th root of $10$ is greater than $1$. This means that $10^x$ is greater than $1$ because the product of $p$ numbers that are greater than $1$ is greater than $1$.
There are no hiccups caused when $x$ is irrational, because in that case $10^x$ is defined in such a way that it is strictly increasing.
