Let $c\in\mathbb{C}$ such that $|c|>e$.
How many solutions are there to the equation $cz^n=e^z$ in $G=\{z\in\mathbb{C}:\Re(z)<1\}$ ?
Equivalently, we can look at zeroes of the function $f(z)=cz^n-e^z$.
Observe that if we look at $D=\{z\in\mathbb{C}:|z|<1\}$ and define $g(z)=cz^n$, then $\forall z\in\partial D$ we have
$$|f(z)-g(z)|=|e^z|=|e^{x+iy}|=|e^x|\leq e<|c|=|g(z)|$$
Then, from Rouché's theorem we obtain $Z_f=Z_g=n$ in $D$.
Is it enough to say that $f$ has only $n$ zeros in $G$ ?
Another observation is that if $w$ is a zero of $f$ of multiplicity $m>1$ then
$$f(w)=cw^n-e^w=0,\ f'(w)=cnw^{n-1}-e^w=0$$
Therefore, $cnw^{n-1}-cw^n=0\implies w^{n-1}(n-w)=0\implies w=0\ \lor\ w=n$.
Now, $w=0$ is impossible since $f(0)=-1\neq 0$ and $w=n$ is not in $G$. We deduce that every zero of $f$ is of multiplicity $1$.
Your proof that every zero of $f$ is of multiplicity $1$ is O.K.
Let $x_0 \in \mathbb R$ and $x_0 \le 0$. Put $r=1-x_0$ and $K(x_0,r):=\{z \in \mathbb C: |z-x_0|<r\}$ (draw a picture !). Then it is easy to see that $0 \in K(x_0,r)$ and $K(x_0,r) \subseteq\{z\in\mathbb{C}:\Re(z)<1\}.$
Then we have $|f(z)-g(z)|=|e^z|=|e^{x+iy}|=|e^x|\leq e<|c|=|g(z)|$ for all $z \in \partial K(x_0,r).$
Therefore we have that for every(!) choice of $x_0 \le 0$, the function $f$ has $n$ zeros in $K(x_0,r).$
Conclusion ?