Fermat's Last Theorem with negative exponent FLT says that the Diophantine equation $a^n+b^n=c^n$ isn't satisfied by any triplet $(a,b,c)$ where $n\in\mathbb{N}$ and $n>2$.
But what happens if $n\in\mathbb{Z}$ and thus can be negative? 
$\textbf{My first thoughts:}$ If we have a negative exponent coupled with a real $a^s$ we can re-write it as $1/a^{-s}.$ So if $n\in\mathbb{Z}$ then the FLT equation takes the rubbish form $$\frac{1}{a^n}+\frac{1}{b^n}=\frac{1}{c^n}$$
which means $$\frac{a^n+b^n}{(ab)^n}=\frac1{c^n}$$
$$\frac{(ab)^n}{a^n+b^n}=\frac{c^n}1$$
$$(ab)^n=(ac)^n+(bc)^n$$
$ab$, $ac$ and $bc$ are all elements of the integer set commonly denoted as $\mathbb{N}$, we thus arrive at a Diophantine equation that is equivalent to the first one we've seen. Therefore the Diophantine equation $a^n+b^n=c^n$ isn't satisfied by any triplet $(a,b,c)$ where $n\in\mathbb{Z}$ and $\mathbf{\color{red}{|n|>2}}$ where $|x|$ is the absolute value of the number $x$.
Is my proof correct?
Furthermore, what happens when the triplet $(a,b,c)\in\mathbb{Z}^3$?

If you had $a^{-5}+b^{-5}=c^{-5}$
then you would also have $(\frac{d}{a})^{5}+(\frac{d}{b})^{5}=(\frac{d}{c})^{5}$ for any $d$
but if we choose $d = lcm(a,b,c)$ then we have a solution to $a^5+b^5=c^5$ in integers.
So we can only solve for $-n$, if we can solve for $n$.
