This is an exercise form Stein-Shakarchi's Complex Analysis (page 155) Chapter 5, Exercise 13:
Prove that $f(z) = e^{z}-z$ has infinite many zeros in $\mathbb{C}$.
Attempt:
If not, by Hadamard's theorem we obtain $$e^{z}-z = e^{az+b}\prod_{1}^{n}(1-\frac{z}{z_{i}})$$ where $\{z_{i}\}$ are the zeros of $f$. How can we conclude ?
Note that $a=1$. Then rewrite your equation as
$$z=e^z P(z)$$
where $P(z)$ is a polynomial of degree $n$. For $|z|$ large enough, $P$ grows of order $|z|^n$. Therefore this equation implies upon taking absolut values that $e^{\mathrm{Re} z}$ decreases like $|z|^{1-n}$ as $|z|\rightarrow\infty$. This is clearly a contradiction.
Therefore you cannot apply Hadamard's theorem, i.e. $f$ has infinitely many zeros.
It is easy to see that $e^z-z$ has order of growth 1. So we can use Hadamard's theorem.
Assume that $f(z)=e^z-z$ has only finitely many zeroes, $a_1,a_2,\ldots, a_n$. Then by Hadamard's factorization theorem, for some $a\in \mathbb{C}$ we have $$e^z-z=e^{az}\prod_{i=1}^{n}\Big(1-\frac{z}{a_i}\Big).$$ Then by using the identity $$\frac{(f_1f_2\ldots f_n)'}{f_1f_2\ldots f_n}=\frac{f_1'}{f_1}+\cdots+\frac{f_n'}{f_n}.$$ We have for $z\notin \{a_1,a_2,\ldots,a_n\}$ $$\frac{e^z-1}{e^z-z}=a+\sum_{i=1}^{n}\frac{1}{z-a_i}.$$
The L.H.S has infinitely many zeroes $2\pi ni$ for $n \in \mathbb{Z}$, but the R.H.S is a quotient of two polynomial functions therefore has only has finitely many zeroes. Therefore $f$ must have infinitely many zeores.
Are you allowed to use Picard's Theorem?
If yes here is a relative question:
Use Picard's Theorem to prove infinite zeros for $\exp(z)+Q(z)$
