$a \leq b$, is $a^b \leq b^a$ correct?

Here is a slightly more systematic way of looking at it. You want to check whether $$a \le b \implies \frac{\log a}{a} \le \frac{\log b}{b}$$ This would be true if the function $f: \mathbb{R} \to \mathbb{R}$ mapping $x \mapsto \frac{\log x}{x}$ were an increasing function.

Now, if you know a little calculus, we can look at the derivative to get a bit of a clue. It's hopefully easy for you to check that $$f'(x) = \frac{1-\log x}{x^{2}}.$$

Now the relationship between increasing/decreasing functions and their derivatives is the sign of the derivative. It is also hopefully easy to see that $$f'(x) \text{ is } \begin{cases} \text{positive } & \text{ if } x < e \\ \text{zero } & \text{ if } x = e \\ \text{negative } & \text{ if } x > e \end{cases}$$

So, $f$ is increasing, stationary, and decreasing on the respective intervals. Oh no! If $f$ is decreasing, then our desired property will fail. So pick some $e < a < b$, and you should get a counterexample. Indeed, $$3 \le 4 \text{ but } 3^{4} > 4^{3}$$ as (un)desired.

Counter-example:

$2\le 5$, yet $\;2^5>5^2$.