A question about exponential functions: $a^b>b^a$ for $b>a>e$

Question

How can we prove that for $b>a>e$ ($e$ being the Euler’s number), $a^b$ is greater than $b^a$?

Answer 1

You want to prove $x^{1/x}$, or equivalently its logarithm $x^{-1}\ln x$, decreases in $x\ge e$ so that $a^{1/a}>b^{1/b}$. Use the product rule.