# 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?

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 $$r<0$$.