Let $k \in(0,\infty)$ , is the function $e^{-|x|^k}$ real analytic for any $(a,b)$ such that $a>0$. 

To show analyticity on an interval $(a,b)$ we need to show that there exists a $C>0$ such that 
\begin{align}
\left| \frac{d^m}{dx^m} e^{-|x|^k} \right| \le C^{m+1} m!, \forall x\in (a,b)
\end{align}

I think, for $|x|>a>0$ can also show the bound of the form

\begin{align}
\left| \frac{d^m}{dx^m} e^{-|x|^k} \right| \le C^{m+1} |x|^{2(k-1)}e^{-|x|^k}
\end{align}

which would imply the analyticity? But I am not sure how to do that. 

**Edit:** By composition theorem, we have that $e^{-|x|^k}$ is analytic on any interval away from zero. See answers below. 

However, I would like to get a proof by showing that a bound of the type holds $\left| \frac{d^m}{dx^m} e^{-|x|^k} \right| \le C^{m+1} |x|^{2(k-1)}e^{-|x|^k}$.