Given $a>b>2$ both positive integers, which of $a^b$ and $b^a$ is larger? Given $a>b>2$ both positive integers, which of $a^b$ and $b^a$ is larger?
I tried an induction approach. First I showed that if $b=3$ then any $a \geq4$ satisfied $a^b<b^a$.
Then using that as my base case I tried to show that given any pair of positive integers $a,b$ satisfying $a>b>2$ and $a^b<b^a$, then $(a+1)^{b+1}<(b+1)^{a+1}$ - but that is where I got stuck.
Any help would be appreciated.
 A: 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 $a^b<b^a$.
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.
A: 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.
A: Another elementary approach:
First notice that it suffices to prove that $n^{n+1}>(n+1)^n$ for $n\ge 3$. We divide both sides by $n^n$ to turn it into $n>(1+1/n)^n$. As a last step, show by induction that $(1+1/n)^n\le n$ for $n\ge3$.  

The case $n=3$ is clear. Now, if $(1+1/k)^k\le k$, then $(1+1/(k+1))^{k+1}\le (1+1/(k))^{k+1}\le k(1+1/(k))\le k+1$.   

Q.E.D.
Inform me if anything needs improvements. Thanks in advance.  
A: Most of the answers submitted are fabulous, but let's endeavor to make this as simple as humanly possible, leveraging intuition primarily.
~Begin by asking yourself: Which is larger, $2^3$ or $3^2$?  $~8<9$. 
~Continue by asking: Which is larger, $3^4$ or $4^3$? $~81>64$. --> This is a reversal of the inequality because the 'weight' of the exponent operation begins to dominate as numbers becomes larger. 
~Conclude by asking: Which is larger, $4^5$ or $5^4$?  $~1024>625$. --> The disparity is more extreme in absolute terms and likely grows as n increases to ∞.
---The trend suggests strongly that $a>b>2$. Of course, more rigor is needed to definitively confirm that the trend does not reverse, but intuition leaves little doubt.
