Question: I was playing a bit whit exponents. I maybe found a working formula for calculating $n^y$ if you know $n^x$. The formula may already be discovered, but the formula I found is:
$$ (n^x)^\frac{y}{x} = n^y $$
Ok, so the formula should work if $n,x,y \in \mathbb N$

I am not sure if it does work whit negative and decimal numbers tho.

Ok, my questions are:  

 1. Can the formula be used whit decimal, and negative numbers (x,y,n)  

 2. Can you prove that the formula works if $x,y,n \in \mathbb N$, if it is possible also for $x,y,n \in \mathbb Q$, $x,y,n \in \mathbb R$ and $x,y,n \in \mathbb Z$.

Answer: 
Let $n=-1$, $x=2$ and $y=1$. Then you have

$$((-1)^2)^{1/2} = 1^{1/2} = 1$$

but 

$$(-1)^1 = -1.$$

So your rule fails.
That is one of the exponent rules.

The rule is: $(x^a)^b = x^{ab}$, But generally, you can't get $n^y$ if you have $n^x$, per example if you have $7^{12}$, you can't get $7^{20} = 7^{12} \cdot 7^8$, without calculating $7^8$.