Can you prove this product equal to sum identity involving exponentiation of the sum? Can you prove this identity?

$$\exp \left(\sum _{n=1}^{\infty} \frac{1}{n \left(1-c^n\right)}\right)=\prod _{n=1}^{\infty} \left(1-\frac{1}{c^n}\right)$$

where $|c|&gt;2$.

I found this conjecture by entering the expression similar to the expression in the left hand side:
$$\sum _{n=1}^{\infty} \frac{1}{n \left(1-2^n\right)}$$ into Excel and looking up the number in the advanced version of the inverse symbolic calculator which told me that:

$$\exp \left(\sum _{n=1}^{\infty} \frac{1}{n \left(1-2^n\right)}\right)=\prod _{n=1}^{\infty} \left(1-\frac{1}{2^n}\right)$$

After that I generalized by guessing that $2$ could be replaced by $c$.

Update 25.1.2017:

Let 
$$s=0 \text{ or } s=1$$
$$\sum _{n=1}^{\infty} \frac{1}{\left(1-c^n\right) n^s}=\sum _{n=1}^{\infty} \lim_{z\to s} \, (z-2) \zeta (z) \left(1-\frac{1}{\left(\frac{c^n}{c^n-1}\right)^{z-1}}\right)$$

Associated Mathematica code of these two small attempts to generalize as above:

(*Mathematica start*)
Clear[s, z, n]
s = 0;
c = N[2, 50];
Sum[1/(n^s*(1 - c^n)), {n, 1, 200}]
Sum[Limit[(z - 2)*Zeta[z]*(1 - 1/(c^n/(c^n - 1))^(z - 1)), 
  z -&gt; s], {n, 1, 100}]

s = 1;
c = N[2, 50];
Sum[1/(n^s*(1 - c^n)), {n, 1, 200}]
Sum[Limit[(z - 2)*Zeta[z]*(1 - 1/(c^n/(c^n - 1))^(z - 1)), 
  z -&gt; s], {n, 1, 100}]
(*end*)


-1.6066951524152917637833015231909245804805796715058

-1.606695152415291763783301523190135719575358660

-1.2420620948124149457978454818946296689734039782504

-1.242062094812414945797845481893840808068182966
Possible hint.

$$e^{\sum_{n=1}^{\infty}\frac{1}{n \left(1-c^n\right)}} = e^{\frac{1}{1\cdot \left(1-c^1\right)} + \frac{1}{2\cdot \left(1-c^2\right)} + \frac{1}{3\cdot \left(1-c^3\right)} + \ldots} = e^{\frac{1}{1\cdot \left(1-c^1\right)}} e^{\frac{1}{2\cdot \left(1-c^2\right)}} e^{\frac{1}{3\cdot \left(1-c^3\right)}} e^{\ldots} = \prod_n e^{\frac{1}{n\cdot \left(1-c^n\right)}}$$
Possible hint.

$$e^{\sum_{n=1}^{\infty}\frac{1}{n \left(1-c^n\right)}} = e^{\frac{1}{1\cdot \left(1-c^1\right)} + \frac{1}{2\cdot \left(1-c^2\right)} + \frac{1}{3\cdot \left(1-c^3\right)} + \ldots} = e^{\frac{1}{1\cdot \left(1-c^1\right)}} e^{\frac{1}{2\cdot \left(1-c^2\right)}} e^{\frac{1}{3\cdot \left(1-c^3\right)}} e^{\ldots} = \prod_n e^{\frac{1}{n\cdot \left(1-c^n\right)}}$$