Given the equation $cz^n-e^z=0$, $|c|>e , n\in \mathbb{N} $.
Show that all the roots of $f(z)= cz^n-e^z$ in $\bar D= \{|z|\leq 1\}$ are from multiplicity of 1.
I used Rouché's theorem by defining $g(z)=cz^n$ to show that $f(z)$ has $n$ roots inside the unit disk.
If I could prove that for every root $w$, s.t $f(w)=0$, we get that $f'(w)\neq 0$, it would solve the rest, but I'm not sure how to do that.
Would appreciate any help:)
Note that $f'(z)=cnz^{n-1}-e^z$ and the if $f(z)=f'(z)=0$, then $cz^n=cnz^{n-1}$ and therefore $z=n$ or $z=0$. But neither $n$ nor $0$ are roots of $f$.