# $f(z)$ has infinitely many zeros and that each zero is simple.

Let $f(z)=e^z-z$

I want to check $f(z)$ is finite order.

And how to show that $f(z)$ has infinitely many zeros and that each zero is simple.

Dfn: an entire function f is finite order if $\exists$ integer $k \gt 0$ and $R\gt 0$ such that for all $|z|\ge R$, we have $|f(z)| \le e^{|z|^k}$

Theorem: suppose $f$ is an entire function of finite order. Then either f has infinite many zeros or $f(z)=Q(z)e^{P(z)}$ where $Q$ and $P$ are polynomials.

Well, from the definition it follows that $f(z)$ is of order $1.$ (since $z$ is of order $1,$ and you should show that the sum of two functions of order $1$ is of order $1.$)
For the second, to show that $f$ has infinitely many zeros, you need to show that it is not of the form $Q(z) \exp(P(z)).$ If it were, $P(z)$ would have to be a linear polynomial (by comparing order) of form $P(z) = a + z.$ It should be easy to check that this is not possible -- if it were, $z$ would grow exponentially).
For the last question, a zero is multiple if it is also a zero of the derivative. The derivative of $f$ is $\exp z - 1,$ so $z$ is a zero of both the function and the derivative if $\exp z - z = \exp z - 1 = 0.$ Is that possible?