Is the function $e^{-|x|^k}$ analytic on any interval not containing zero?

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

Solutions Collecting From Web of "Is the function $e^{-|x|^k}$ analytic on any interval not containing zero?"

Yes: If $a < x < b$, then $|x| = x$, and $|x|^{k} = x^{k}$ is real-analytic (Newton’s binomial theorem), and a composition of real-analytic functions is real-analytic.

The requested bound cannot hold.

From

$$\left(e^P\right)’=P’e^P,\\\left(Qe^P\right)’=(Q’+QP’)e^P=Re^P$$

we deduce that the degree of the polynomial factor in front of the exponential follows the recurrence

$$\begin{cases}q=0\to r=p-1,\\q>0\to r=pq-1.\end{cases}$$

Then with $P=-x^k$, the degrees are $0,k-1,k^2-k-1,k^3-k^2-k-1,\cdots k^m-\frac{k^m-1}{k-1}$, and grow much larger than $2(k-1)$.