Given $a>b>2$ both positive integers, which of $a^b$ and $b^a$ is larger?

Suppose for a moment: $$a^b = b^a.$$ Taking natural log: $$ b\ln a = a\ln b, $$ which is $$ \frac{\ln a}{a} = \frac{\ln b}{b} . $$ Now consider the function: $$ f(x) = \frac{\ln x}{x}, $$ where $$ f'(x) = \frac{1 - \ln x}{x^2} < 0 \quad \text{ if }\;x>e. $$ I believe you could take it from here.

The result follows easily using calculus. Here's an elementary approach, which uses the fact that $a, b$ are positive integers.

Consider $n \geq 3$. Then $$(n+1)^n=\sum_{i=0}^{n}{\binom{n}{i}n^{n-i}}=1+n^2+\sum_{i=0}^{n-2}{\binom{n}{i}n^{n-i}}<n^n+\sum_{i=0}^{n-2}{n^n}=n^{n+1}$$ since $\binom{n}{i} \leq n^i$, and $1+n^2<n^n$ for $n \geq 3$.

Therefore $n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}$ for $n \geq 3$. This immediately implies that $a^{\frac{1}{a}}<b^{\frac{1}{b}}$, so $a^b<b^a$.

Hint:

Which is larger, $b\ln a$ or $a\ln b$?

Which is larger, $\dfrac{\ln a}{a}$ or $\dfrac{\ln b}{b}$?

How does $\dfrac{\ln x}{x}$ behave as $x$ increases? Looks like a job for the derivative.