Is $z^{-1}(e^z-1)$ surjective? Is the entire function
$$
   f(z)=\frac{1}{z}(e^z-1)
$$
surjective?
I tried to argue like in the question below, but it does not seem to work in a similar way. 
$z\exp(z)$ surjectivity with the Little Picard Theorem
 A: The Little Picard Theorem is a bit strong for this problem and the one you linked to. The good thing about these two problems is that the functions involved are of finite growth. Here, $f$ has order $1$.
If $w \not \in im(f)$ for some $w \in \mathbb{C}$, then $f(z)-w$ is an entire function of order $1$ with no zeroes. Therefore, Hadamard's factorization theorem implies $f(z)-w = e^{az+b}$ for some $a,b \in \mathbb{C}$. This yields $e^z-1-zw = ze^{az+b}$ for all $z \not = 0$. This doesn't look possible. I'll let you finish from here. 
A: Since $f(z)=\frac{e^z-1}{z}$ is entire and non-constant, by Little Picard's theorem it is either surjective or just misses a single complex value. Let us assume to be in the second case. The missing value has to be a real value by Schwarz' reflection principle, since $f$ is real over the real line (assuming that $f$ misses some $w\in\mathbb{C}\setminus\mathbb{R}$, it also misses $\overline{w}$). The missing value is not zero since $f(2\pi i)=0$. The missing value is not some $\alpha>1$ since the line $y=1+\alpha z$ meets the graph of $g(z)=e^z$ at some point with a positive abscissa by the convexity of $g(z)$. The missing value is not some $\alpha\in(0,1)$ since the line $y=1+\alpha z$ meets the graph of $g(z)=e^z$ at some point with a negative abscissa by the convexity of $g(z)$. Negative values are also attained, hence it follows that the only missing value can be $1$, but $1=f(0)$, so $f$ is surjective.
