Deduce from Hadamard's Theorem that if $F$ is entire and has non-integral order of growth then it has infinitely many zeros. Deduce from Hadamard's Factorization Theorem that if $F$ is entire and has finite non-integral order of growth then it has infinitely many zeros. This is exercise 14 of chapter 5 in Stein and Shakarchi's Complex Analysis. 
Hadamard's theorem states that if $F$ is entire and has growth order $\rho_0$, and $k=\lfloor \rho_0 \rfloor$, $a_1,a_2,\ldots$ are the zeros of $F$ then 
$$ F(z)=e^{P(z)}z^m\prod_{n=1}^\infty E_k(z/a_n),$$
where $P$ is a polynomial of degree at most $k$ and $m$ is the order of the zero of $f$ at $z=0$. 
How can I use the theorem? Assume that $F$ has finitely many zeros and get a contradiction perhaps?
 A: EDIT: The old answer is wrong. $f=f_1f_2$ where $f_1(z)=e^z$ and $f_2(z)=e^{-z}$ is a counter-example.
Suppose $f$ has only finitely many zeroes in $\mathbb{C}$, say, $\{a_1,...,a_N\}$. Let $g(z)=\prod_{n=1}^N (z-a_n) \,\forall z \in \mathbb{C}.$ Then $\frac{f}{g}$ has no zeroes in $\mathbb{C}.$
Now apply Hadamard factorizaton: $\frac{f(z)}{g(z)}=e^{P(z)} \,\forall z \in \mathbb{C}$, where $P(z)$ is a polynomial. Hence $f(z)=g(z)e^{P(z)} \forall z \in \mathbb{C}.$ 
Since $g$ is a polynomial of degree $N$, the order of growth of $f$ equals $\deg(P)$. By the infimum property it cannot be larger, for, if $f=f_1f_2$ where $|f_1(z)| \le A_1e^{B_1|z|^{\rho_1}} \forall z \in \mathbb{C}$ and $|f_2(z)| \le A_2e^{B_2|z|^{\rho_2}} \forall z \in \mathbb{C}$, we must have $|f(z)|=|f_1(z)f_2(z)| \le A_1A_2e^{(B_1+B_2)|z|^{max\{\rho_1,\rho_2\}}} \forall z \in \mathbb{C}$. If it were smaller, say, $\rho_0$, taking $z \to \infty, z \in \mathbb{R}$, you get a contradiction, since $|g(x)| \le Ae^{{B|x|^{\rho_0}-P(x)}} \le A'e^{B'|x|^r}$ for sufficiently large $x$, for some $r<0$.
