How does $a^n$ compare to $b^n$, in case $a>b$ and $n>0$? ( On an inequality rule regarding exponentiation). Suppose $a>b$ and that $n > 0$. 
How does $a^n$ compare to $b^n$? 
By considering various cases , I arrived at the following rule : 
"In case $a> b$ , and $n > 0$ , then $n$-powers conserve the order of $a$ and $b$ , that is ,$a^n > b^n$ , except when n is even and  


*

*either  (1) $a$ and $b$ are both negative  , 

*or (2) $a$ and $b$ have different signs  and $a$ is smaller than $b$ in absolute value." 
Is this rule corrrect? In case it is, is there a simpler way to phrase it? 


Yes this rule is true : 
$f(x)=x^n$ then $f'(x)=nx^{n-1}$ for positive value the derivative is positive and therefore the function is increasing. 
for negative value : 
if n is even n-1 is odd and therefore for negative value the function is decreasing 
if n is odd n-1 is even and therefore the function is increasing.
