An infinite product for $\left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^2$? Given the Dedekind eta function,
$$\eta(\tau) = q^{1/24} \prod_{n=1}^\infty (1-q^n)$$
where $q = \exp(2\pi i\tau)$.  Consider the following "family",
$\begin{align}
\left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{24} &= \frac{u^8}{(-1+16u^8)^2},\;\;\; u = q^{1/8} \prod_{n=1}^\infty \frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})^2}\\[2.5mm]
\left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{12} &= \frac{c^3}{(1+c^3)(-1+8c^3)^2},\;\;c = q^{1/3} \prod_{n=1}^\infty \frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})^2}\\[2.5mm]
\left(\frac{\eta(5\tau)}{\eta(\tau)}\right)^{6}\; &= \frac{r^5}{(r^5+u_5^5)(r^5-u_5^{-5})},\quad r\; =\; q^{1/5} \prod_{n=1}^\infty \frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}\\[2.5mm]
\left(\frac{\eta(7\tau)}{\eta(\tau)}\right)^{4}\;&= \frac{h(h-1)}{1+5h-8h^2+h^3},\quad h = 1/q\, \prod_{n=1}^\infty \frac{(1-q^{7n-2})^2(1-q^{7n-5})^2(1-q^{7n-3})(1-q^{7n-4})}{(1-q^{7n-1})^3(1-q^{7n-6})^3}\\[2.5mm]
\left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^{2} &=\frac{s}{(s-u_{13})(s+u_{13}^{-1})},\quad\; s =\; ???\\
\end{align}$
with fundamental units $u_n$ as $u_5 = \frac{1+\sqrt{5}}{2}$ and $u_{13} = \frac{3+\sqrt{13}}{2}$. The second-to-the-last appears in Chap 10 (10.2) of Duke's Continued Fractions and Modular Functions. 
Question: What is the analogous infinite product, if any, for $\left(\frac{\eta\,(13\tau)}{\eta\,(\tau)}\right)^2$ similar to the ones above? 
Postscript: This question has been modified before, as it was a bit unclear.

Michael Somos just today found the identity,
$$
\left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^{2} = \frac{s}{s^2-3s-1}$$
where,
$$s=\frac{1}{q}\; \prod_{n=1}^\infty \frac{ (1-q^{13n-2})(1-q^{13n-5})(1-q^{13n-6})(1-q^{13n-7})(1-q^{13n-8})(1-q^{13n-11}) }{(1-q^{13n-1})(1-q^{13n-3})(1-q^{13n-4})(1-q^{13n-9})(1-q^{13n-10})(1-q^{13n-12})} $$
thus completing the family for $N = 2,3,5,7,13$.
Does this address Matt E and Loeffler's comments? Is "s" a modular function? 
