While comparing the $123^{124}$ and $124^{123}$ (just for fun) I come up with an interesting question.
Is it possible to find the least possible natural $n$, such that $a = b - k, k > 0 \quad and \quad a^n > b^{n-1}$ ? $a$ , $b$ and $k$ are fixed.
I have no idea how to approach the problem.
Thank you.
As lulu suggested, take the logarithms and use the logarithm law $$\log(a^n)=n\cdot \log(a)$$ You get an inequality that you can easily solve for $n$.