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I would ask you for some hints for the calculation of the product $$\prod_{n=1}^\infty(1-e^{-18 n\pi}).$$

Is it possible to be approached without special functions?

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You need to go step by step. Let $$F(q) =\prod_{n=1}^{\infty} (1-q^{2n})\tag{1}$$ where $0<q<1$. Then you wish to seek the value of $F(q^9)$ where $q=e^{-\pi} $. Fortunately Ramanujan gave relations between $F(q), F(q^3)$ and $F(q^9)$ so that $F(q^9)$ can be evaluated in closed form if the value of $F(q) $ is known.

The value of $F(q) $ for $q=e^{-\pi} $ is well known and can be obtained using the link between such functions and elliptic integrals. Thus if $$\eta(q) =q^{1/12}F(q)\tag{2}$$ and $k$ is the elliptic modulus corresponding to nome $q$ and $K$ the corresponding complete elliptic integral then we have $$\eta(q) =2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{3}$$ If $q=e^{-\pi} $ then $k=k'=1/\sqrt{2}$ and $K=\Gamma^2(1/4)/4\sqrt{\pi}$ and thus $$\eta(q) =\frac{\Gamma(1/4)}{2\pi^{3/4}}$$ and hence $$F(q) =e^{\pi/12}\cdot \frac{\Gamma(1/4)}{2\pi^{3/4}}\tag{4}$$ If $L, l $ correspond to $q^3$ then we have $$\eta(q^3)=2^{-1/3}\sqrt{\frac{2L}{\pi}}(ll')^{1/6}$$ From this answer the values $L, l, l'$ are known and we have thus $$\eta(q^{3})=\frac{\Gamma(1/4)}{\pi^{3/4}}\frac{\sqrt[4]{3+2\sqrt{3}}\sqrt[3]{2-\sqrt{3}}}{2\sqrt{3}}$$ You should check that the above expression matches with the value given in Wikipedia for $\eta(3i)$ and we thus have $$\eta(q^3)=\frac{\Gamma (1/4)}{2\sqrt[3]{3}\sqrt[12]{3+2\sqrt{3}}\pi^{3/4}}$$ so that $$F(q^3)=e^{\pi/4}\cdot \frac{\Gamma (1/4)}{2\sqrt[3]{3}\sqrt[12]{3+2\sqrt{3}}\pi^{3/4}}$$ Now we need to use Ramanujan's identity $$1+9q^2\cdot\frac{F^3(q^9)}{F^3(q)}=\left(1+27q^2\cdot\frac{F^{12}(q^3)}{F^{12}(q)}\right) ^{1/3}\tag{5}$$ and we can get the value $F(q^9)$ in closed form.

We have $$1+27q^2\frac {F^{12}(q^3)}{F^{12}(q)}=\frac{2\sqrt{3}+6}{9}$$ and hence $$F(q^9)=e^{3\pi/4}\cdot\frac{\sqrt[3]{\sqrt[3]{18+6\sqrt{3}}-3}}{6}\cdot\frac{\Gamma (1/4)}{\pi^{3/4}}$$ Note that the first term of your product is practically equal to $1$ and the above complicated closed form expression is thus approximately equal to $1$. To be more precise equating the above expression with $1$ gives the value of $\Gamma(1/4)$ correct to 23 places of decimals as $$\Gamma(1/4)=3.625609908221908311930686156\dots$$ whereas the correct value is $$\Gamma(1/4)=3.625609908221908311930685155\dots$$

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Not yet a complete answer

$$P=\prod_{n=1}^{\infty}(1-e^{-18\pi n})$$ I don't think it is possible without special functions, but it is worthwhile to recall the definition of the Euler function $$\phi(q)=\prod_{n=1}^{\infty}(1-q^n)$$ Immediately we recognize the product in question as $$P=\phi(e^{-18\pi})$$ Which may help, because Ramanujan found that $$\phi(e^{-\pi})=\frac{e^{\pi/34}}{2^{7/8}}Y$$ $$\phi(e^{-2\pi})=\frac{e^{\pi/12}}{2}Y$$ $$\phi(e^{-4\pi})=\frac{e^{\pi/6}}{2^{11/8}}Y$$ $$\phi(e^{-8\pi})=\frac{e^{\pi/3}}{2^{29/16}}(\sqrt{2}-1)^{1/4}Y$$ Where $$Y=\frac{\Gamma(1/4)}{\pi^{3/4}}$$ There may be a pattern to these closed forms, so there could still be hope...

Also we may see that $$ \begin{align} \log P=&\sum_{n=1}^{\infty}\log(1-e^{-18\pi n})\\ =&\sum_{n=1}^{\infty}\log\bigg(\frac{e^{18\pi n}-1}{e^{18\pi n}}\bigg)\\ =&\sum_{n=1}^{\infty}\log(e^{18\pi n}-1)-18\pi n\\ \end{align} $$ Which may or may not help

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