Is my inequality correct? And how do I prove it?

After watching many videos about "which number is bigger?", I decided to make the following inequality: $$b^a<a^b$$ where $b>a>e$.

Is this inequality correct? If it's not correct, how can I modify it to make it correct? And if it's correct, how can I prove it?

Edit: Please tell me what's wrong with my question.

Taking logarithm, we get the equivalence

$$b^a<a^b\iff a\ln(b)<b\ln(a)$$ $$\iff \frac{\ln(b)}{b}<\frac{\ln(a)}{a}$$

For $ x>e $, considere $$f(x)=\frac{\ln(x)}{x}$$

$ f $ is differentiable at $ [e,\infty) $ and $$f'(x)=\frac{1-\ln(x)}{x^2}<0$$ since $ \ln(x)>\ln(e)$.

Thus

$$b>a\implies f(b)<f(a)$$ $$\implies b^a<a^b$$