I am looking to determine the order of $f(z) = \exp Q(z)$ when $Q$ is a polynomial of degree $q$.
I think the order is $q$, but I am struggling to prove it. The definition of order is:
An entire function $f$ is said to be of order $\rho$, if $$ |f(z)| < \exp r^{\rho+\varepsilon}, |z|=r$$ for sufficiently large values of $r$ and $$|f(z)| > \exp r^{\rho-\varepsilon}, |z| = r$$ for infinitely many, arbitrarily large values of $r$, where $\varepsilon$ denotes an arbitrary positive number.
Indeed, the order is $q$. Write $Q(z):=\sum_{j=0}^qc_jz^j$, where $c_q\neq 0$. We have for $|z| =r$: $$\tag{*} |f(z)|\leqslant \exp\left(\sum_{j=0}^q|c_j|r^j\right),$$ and since $$\lim_{r\to \infty} \frac 1{r^{q+\varepsilon}} \sum_{j=0}^q|c_j|r^j=0,$$ we have $\sum_{j=0}^q|c_j|r^j\lt r^{q+\varepsilon} $ for $r$ large enough, hence by (*), it follows that $|f(z)|\lt \exp(r^{q+\varepsilon})$.
Now we check the second inequality. This time, we write for $|z|=r$, $$\tag{**}\frac{|f(z)|}{\exp(r^{q-\varepsilon} )}=\left|\exp\left(\sum_{j=0}^qc_jz^j-r^{q-\varepsilon}\right)\right| , $$
and $$\tag{***} \sum_{j=0}^qc_jz^j-r^{q-\varepsilon}=r^q\left(c_q\frac{z^q} {r^q} +\sum_{j=0}^{q-1}c_jz^{j}r^{-q} + r^{-\varepsilon}.\right)$$
We choose a sequence $(z_n)_{n\geqslant 1}$ such that $|z_n| =n$ and $c_qz_n^q=n^q|c_q|$. By () and (*), we have $$\frac{|f(z_n)|}{\exp(n^{q-\varepsilon} )}\geqslant \left|\exp\left( n^q\left(|c_q| +\delta_n\right)\right)\right|,$$ where $\delta_n\to 0$.