I am reading Stein and Shakarchi's Complex Analysis and came across this question which seems to be implicit in the text. First some definitions.
An entire function has an order of growth $\le \rho$ if $$|f(z)| \le Ae^{B|z|^\rho},$$ and the order of growth $\rho_0$ of $f$ is the infimum of all such $\rho$'s.
In this case, it seems like we have for any $s>\rho_0$, $f$ has an order of growth $\le s$.
How do we prove this? $\rho_0$ being the infimum of all such $\rho's$ means that we have infinitely many $\rho_i$ close to $\rho_0$ such that $|f(z)| \le Ae^{B|z|^{\rho_i}}$. But I can't figure out how to get an upper bound by a power of $s$ in the exponent. I would greatly appreciate any help.
To be more specific about the context, this question follows from here where $E(z)=z^m \Pi_{n=1}^\infty E_k(z/a_n)$.
So the lower bound for $E$ is given in Corollary 5.4, which gives the bound
so using this bound and the fact that $|z|=r_m$ for $r_m $ growing to $\infty$, we can bound $|z|^m$ above by $1$ as well and be left with $e^{c |z|^s}$. Now we have to bound $f$ above by an order of $e^{c |z|^s}$ and this seems to be done from the statement I am asking about.
Note that for any $C,\delta>0$, we have
$$ r^C\leq r^{C+\delta} $$ for all $r\geq 1$.
Furthermore, for $B>0$, we have that $\exp( Br^C)\le \exp(B)$ for all $r\in [0,1]$ and $\exp(Br^{C+\delta})\ge 1.$ Hence, defining $A'= \exp(B)\ge 1$, we have $$ \exp(Br^C)\le A'\exp(Br^{C+\delta}) $$ and hence, if $|f(z)|\leq A\exp(B|z|^C)$, we get that $$ |f(z)|\le AA' \exp(B|z|^{C+\delta}), $$ which is what you wanted.