A function that is Taylor expandable about $x = a$ can be written as $$ f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots+\frac{f^{(k)}(a)}{k !}(x-a)^{k}+R_{k+1}(x)$$ where $R_{k+1}=\frac{f^{(k+1)}\left(\xi\right)}{(k+1) !}(x-a)^{k+1}$ for some $\xi$ between $a$ an $x$.
I have a Taylor expansion (not shown here) where I have derived that
$$R_{k}<\frac{\frac{\mathrm{d}^ke^{x^m}}{\mathrm{d}x^k}}{k !}x^{k} \,. $$
Unfortunately, in need very large $k$ indices such that calculating the derivatives using a computer $\frac{\mathrm{d}^ke^{x^m}}{\mathrm{d}x^k}$ become a hastle (yes this is a weakness with my Taylor expansion). Accordingly, I want to find an upper bound $M_k$ such that
$$R_{k}<\frac{\frac{\mathrm{d}^ke^{x^m}}{\mathrm{d}x^k}}{k !}x^{k} \leq\frac{M_k}{k !}x^{k} \,. $$
This introduction leads to the problem statement: What is your best upper bound for $\frac{\mathrm{d}^ke^{x^m}}{\mathrm{d}x^k}$?
Details: All of my $m$ values are larger than $1$, typically between $3/2$ and $3$. Furthermore, all of my $x$ values are also positive. I do not need the tightest upper bound.