Let $0<s<1$ and consider the power series $$\sum_{n=0}^{\infty}\frac{r^n}{(n!)^{1/s}}.$$ I need to show that for any given $\epsilon>0$, there exists $R>0$ such that for all $r>R$, $$\sum_{n=0}^{\infty}\frac{r^n}{(n!)^{1/s}}<\exp({r^{s+\epsilon}}).$$
This comes from a complex analysis problem, where I need to prove that the entire function $\sum_{n=0}^{\infty}\frac{z^n}{(n!)^{1/s}}$ has order at most $s$, which, in this context, translates to the above problem.
You can show that the order $\rho$ of an entire function $f$ can be computed as $$ \rho = \limsup_{n\to\infty} \frac{n\log n}{\log(1/|a_n|)} $$ where $a_n$ are the Maclaurin coefficients of $f$.
In your case, you have $a_n = \dfrac{1}{(n!)^{1/s}}$, so $$ \rho = \limsup_{n\to\infty} \frac{s\, n\log n}{\log(n!)} = s $$ (using some standard estimates of $n!$).