Yet another question of which is larger: $70^{71}$ or $71^{70}$. I solved it by observing that $f(x)=\frac{\ln(x)}{x}$ is decreasing for all $x>e$ since $f'(x)=\frac{1-\ln(x)}{x^2}<0$ for all $x>e$. Then we have $$ \begin{align*} \frac{\ln(70)}{70}>\frac{\ln(71)}{71} &\iff 71\ln(70)>70\ln(71)\\ &\iff \ln(70^{71})>\ln(71^{70})\\ &\iff e^{\ln(70^{71})}>e^{\ln(71^{70})}\\ &\iff 70^{71}>71^{70}. \end{align*} $$ I briefly search for similar problems along this idea, but didn't really find much in other ideas about how to solve this. So I open this to the group: how else might this be proved?
EDIT: So this problem is a specific case of the more general problem I apparently duplicated. However, we can extend the duplicate problem by allowing real number inputs as opposed to just integer inputs. Thus $x^{(x+1)}>(x+1)^x$ for all reals $x>e$. This is proved following the exact same proof I gave above with the appropriate substitutions: $$ \begin{align*} \frac{\ln(x)}{x}>\frac{\ln(x+1)}{x+1} &\iff (x+1)\ln(x)>x\ln(x+1)\\ &\iff \ln(x^{(x+1)})>\ln((x+1)^{x})\\ &\iff e^{\ln(x^{(x+1)})}>e^{\ln((x+1)^{x})}\\ &\iff x^{(x+1)}>(x+1)^{x}. \end{align*} $$ Thanks for the input everyone!