If we know that $a^n$ grows unbounded for $1<a$ and that $a^n\to 0$ if $a<1$, I am trying to prove that $\{a^n-n\}$ is not Cauchy if $1<a$.

Intuitively we should be able to show that for large enough $m<n$ we have

$$ |m-a^m-(n-a^n)|=|m-n+a^n-a^m|\ge 1$$

Of course we can choose $n$ large enough that $m<a^n$. Not sure how much this helps.

I tried using the reverse triangle inequality but that didn't seem to help.

I could factor out

$$ \left(\frac 1 {a^n}\right)\left|\frac{m}{a^n}-\frac{n}{a^n}+1-a^{m-n}\right|$$

But I don't see how to see any way to use this. And I can't really think of anything else to do to this. I know that if we had the full power of calculus we could use L'Hospital's Rule, but this is too basic to permit using derivatives. We would also need to know the derivative of exponential functions, and the proof of that I recall uses knowledge of the FTC, so I don't think any line of thinking in this direction could be helpful.