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Taking logarithm gives $$ -\sum_{k=a}^N \log(e^{k \varkappa}-1)= -\frac1\varkappa \sum_{k=a}^N \log\left(e^{k \varkappa}-1\right)\varkappa. $$ The last sum is a Riemann sum of the integral $$ \int_{\varkappa a}^{\varkappa N}\log(e^x-1)\,dx= \text{Li}_2(e^{\varkappa a})- \text{Li}_2(e^{\varkappa N})+ i \pi \varkappa( a-N), $$ where $\text{Li}_2(x)$ is the polylogarithm. So the product is approximately equal to $$ e^{\frac1\varkappa (\text{Li}_2(e^{\varkappa N})- \text{Li}_2(e^{\varkappa a}))+i \pi \varkappa(N-a)}. $$

Taking logarithm gives $$ -\sum_{k=a}^N \log(e^{k \varkappa}-1)= -\frac1\varkappa \sum_{k=a}^N \log\left(e^{k \varkappa}-1\right)\varkappa. $$ The last sum is a Riemann sum of the integral $$ \int_{\varkappa a}^{\varkappa N}\log(e^x-1)\,dx= \text{Li}_2(e^{\varkappa a})- \text{Li}_2(e^{\varkappa N})+ i \pi \varkappa( a-N), $$ where $\text{Li}_2(x)$ is the polylogarithm. So the product is approximately equal to $$ e^{\frac1\varkappa (\text{Li}_2(e^{\varkappa N})- \text{Li}_2(e^{\varkappa a}))+i \pi (N-a)}. $$

Taking logarithm gives $$ -\sum_{k=a}^N \log(e^{k \varkappa}-1)= -\frac1\varkappa \sum_{k=a}^N \log\left(e^{k \varkappa}-1\right)\varkappa. $$ The last sum is a Riemann sum of the integral $$ \int_{\varkappa a}^{\varkappa N}\log(e^x-1)\,dx= \text{Li}_2(e^{\varkappa a})- \text{Li}_2(e^{\varkappa N}), $$ where $\text{Li}_2(x)$ is a polylogarithm function. So the product is approximately equal to $$ e^{\frac1\varkappa (\text{Li}_2(e^{\varkappa N})- \text{Li}_2(e^{\varkappa a}))}. $$

Taking logarithm gives $$ -\sum_{k=a}^N \log(e^{k \varkappa}-1)= -\frac1\varkappa \sum_{k=a}^N \log\left(e^{k \varkappa}-1\right)\varkappa. $$ The last sum is a Riemann sum of the integral $$ \int_{\varkappa a}^{\varkappa N}\log(e^x-1)\,dx= \text{Li}_2(e^{\varkappa a})- \text{Li}_2(e^{\varkappa N})+ i \pi \varkappa( a-N), $$ where $\text{Li}_2(x)$ is the polylogarithm. So the product is approximately equal to $$ e^{\frac1\varkappa (\text{Li}_2(e^{\varkappa N})- \text{Li}_2(e^{\varkappa a}))+i \pi \varkappa(N-a)}. $$

Taking logarithm gives $$ -\sum_{k=a}^N \log(e^{k \varkappa}-1)= -\frac1\varkappa \sum_{k=a}^N \log\left(e^{k \varkappa}-1\right)\varkappa. $$ The last sum is a Riemann sum of the integral $$ \int_{\varkappa a}^{\varkappa N}\log(e^x-1)\,dx= \text{Li}_2(e^{\varkappa a})- \text{Li}_2(e^{\varkappa N}), $$ where $\text{Li}_2(x)$ is a polylogarithm function. So the product is approximately equal to $$ e^{\frac1\varkappa (\text{Li}_2(e^{\varkappa N})- \text{Li}_2(e^{\varkappa a})}. $$

Taking logarithm gives $$ -\sum_{k=a}^N \log(e^{k \varkappa}-1)= -\frac1\varkappa \sum_{k=a}^N \log\left(e^{k \varkappa}-1\right)\varkappa. $$ The last sum is a Riemann sum of the integral $$ \int_{\varkappa a}^{\varkappa N}\log(e^x-1)\,dx= \text{Li}_2(e^{\varkappa a})- \text{Li}_2(e^{\varkappa N}), $$ where $\text{Li}_2(x)$ is a polylogarithm function. So the product is approximately equal to $$ e^{\frac1\varkappa (\text{Li}_2(e^{\varkappa N})- \text{Li}_2(e^{\varkappa a}))}. $$