Is there a way to prove this exponential inequality: if $a>b$ then $a^a>b^b$ for $a,b>1$?

You're asking to prove that the map $f(x):=x^x=\exp(x\log x)$ is strictly increasing. But since $x\mapsto x$ and $x\mapsto\log x$ are both strictly increasing and the first one is $>1$ in $\Bbb R_{>1}$, their product is such (in $\Bbb R_{>1}$).

Then the exponential is strictly increasing too, so $f$ is the composition of strictly increasing maps, thus itself is strictly increasing too.

If $a\gt b$ and $a\gt1$ and $b\gt1$ and $a,b\in\mathbb{R}$ then this implies:$$a^b\gt b^b\tag{1}$$

Also, if $a\gt b$ then this implies:$$a^a\gt a^b\tag{2}$$

Now, making use of result (1) in result (2) yields:$$a^a\gt a^b\gt b^b$$

Note that the derivative of $x^x$ is $(\ln(x)+1)x^x>0$(see here) if $x>e^{-1}$, implying $x^x$ is a strictly increasing function if $x>e^{-1}$.

This implies your claim is true not only when $a,b>1$ but also $a,b>e^{-1}$