If $a>b$ then $a^a>b^a. $ If $a>b$ then $a^a>b^a. $ where $a, b\in \mathbb{R}. $
I am unable to prove this. I have seen proofs of this in case of Integers but what about Real numbers. 
Any help will be appreciated. 
 A: You need $a, b > 0$ , otherwise $a^a$ and $b^a$ are  not defined. Having said that, you need to use some definition of $a^a$; what is your definition?
One could cheat and say that $a^a = e^{a \log a}$, hence 
$$e^{a\log a } > e^{a \log b} \iff a\log a > a \log b \iff \log a > \log b \iff a > b$$
but here we used the fact that both $e^x$ and $\log x$ are increasing functions. Does this satisfy you?
Otherwise you could define first $a^n = a \cdots a$ ($n$ times) for $n$ integer, then generalize this definition to rationals and then to reals in an appropriate way. If you follow these generalizations you will see how the property $a> b \implies a^n > b^n$ is kept throughout. 
Finally, note that $a > b \implies a^x > b^x$ for every $x \in \mathbb R$, not only $x = a$ (as you can prove in the same way as above)
A: This is not true without the restriction that $a, b > 0$.  If we have $a > b > 0$ then we have:
\begin{align}
  a &> b\\
  \ln a &> \ln b\\
  a \ln a &> a \ln b\\
  \ln (a^a) &> \ln (b^a)\\
  a^a &> b^a
\end{align}
A: Let $a > b > 0$. Then
$a^{a} > b^{a}$ iff
$$
\bigg( \frac{a}{b} \bigg)^{a} > 1.
$$
Since $a > b > 0$, we have $x := a/b > 1$; but $y^{a} > 1$ for all $y > 1$, so $x^{a} > 1$ and we are done.
