I know there are $n$ zeros in the open ball $B_1(1)$ from Rouché's Theorem. However, how do I show that the order of these zeros are all one? I don't know where the zeros are, and this should be true for any arbitrary $n$. One thing I want to try is to show that $f'(z_0)\neq 0$ for any zero $z_0$, so we have $$ f'(z) = 2n(z-1)^{n-1} + e^{-z}. $$
Since $e^{-z_0} =2(z_0-1)^n$, this implies $$ 2n(z_0-1)^{n-1} + 2(z_0-1)^n = 2(z_0-1)^{n-1}(n+z_0-1). $$
Now I am stuck on what to do, since I would have to show that neither of the two factors equals $0$. Any hints?