How can we show the other direction? I want to prove the following implication: 
$$k \in \mathbb{Z} \Leftrightarrow ce^x-1 \mid c^ke^{kx}-1$$ 
For the direction $\Rightarrow$ I tried the following: 


*

*$k >0$: 
$$\sum_{i=0}^{k-1} (ce^x)^i=\frac{(ce^x)^k-1}{ce^x-1} \\ \Rightarrow (ce^x)^k-1=(ce^x-1)\sum_{i=0}^{k-1} (ce^x)^i$$ 
So when $k$ is an integer $>0$ we have that $ce^x-1 \mid c^ke^{kx}-1$. 

*$k <0$: 
$$\sum_{i=0}^{k-1} ((ce^x)^{-1})^i=\frac{\left (\frac{1}{ce^x}\right )^k-1}{\frac{1}{ce^x}-1}=\frac{\frac{1-(ce^x)^k}{(ce^x)^k}}{\frac{1-ce^x}{ce^x}}=\frac{ce^x}{(ce^x)^k} \frac{1-(ce^x)^k}{1-ce^x} =\frac{1}{(ce^x)^{k-1}} \frac{(ce^x)^k-1}{ce^x-1} \\ \Rightarrow (ce^x)^k-1=(ce^x-1)(ce^x)^{k-1}\sum_{i=0}^{k-1} ((ce^x)^{-1})^i$$ 
So when $k$ is an integer $<0$ we have that $ce^x-1 \mid c^ke^{kx}-1$. 


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Is this correct? 
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Could you give me a hint how we could show the other direction? 
We suppose that $ce^x-1 \mid c^ke^{kx}-1$, that means the solutions of $ce^x-1$ are also solutions of $c^ke^{kx}-1$, right? 
Does this help? 
 A: Assuming $x$ can take complex values (which makes some sense since the coefficients of the ring of functions, as well as multipliers $\lambda$ on the generators $e^{\lambda x}$ are any complex numbers), we can find the zeros of $c e^x -1$ by solving $e^x=1/c,$ obtaining
$$x = \log(1/a)+2 \pi n i $$
with $n \in \mathbb{Z},$ and the log is chosen in the strip with imaginary part in $[0,2\pi)$. Now if $c e^x-1$ divides $c^k e^{kx}-1$ then the zero set of the first must be a subset of the zero set of the second.
So we can determine the latter zero set by solving $e^{kx}=1/c^k,$ obtaining $kx=\log(1/c^k)+2 \pi m i$ and so that zero set is
$$x = (1/k)\log(1/c^k)+ \frac{2 \pi m i}{k}. \tag{*}$$
For example if $k=1/2$ the second zero set is $(1/2)log(1/c^2)+4 \pi m i,$ which fails to contain the first zero set, because the spacing of the second zero set is $4 \pi i$ between zeros adjacent vertically, as compared to the first zero set whose spacing is $2 \pi i$ between adjacent zeros. It seems fairly clear that if $k$ is not an integer the second zero set will fail to contain the first. [But that needs to be shown, will leave it undone here...]
NOTE: The OP Mary Star had already suggested comparing the zero sets in the post.
Another note: a previous version of this had used the false "law" $\log(kx)=k\log x$ which need not hold for complex $k$ (even if one changes "branches).
