Direct proof of the level four eta identity Of Somos' over 6200 eta identities, the one of smallest level (maybe the simplest?) is the eta identity of level $4$ $$\eta(z)^{16} \eta(4z)^8 + 16 \eta(4z)^{16} \eta(z)^8 = \eta(2z)^{24}.$$
As with all identities among modular forms, we can prove it by checking the first few coefficients. I wonder if it is possible to give a "real" proof of this straight from the product $$\eta(z) = q^{1/24} \prod_{n=1}^{\infty} (1 - q^n)$$ in the style of this answer. The higher exponents in this problem seem to make it harder to use Jacobi's triple product.

This identity
$$\eta(z)^{16} \eta(4z)^8 + 16 \eta(4z)^{16} \eta(z)^8 = \eta(2z)^{24}$$
written as
$$\bigg(\frac{\eta(z)}{\eta(2z)}\bigg)^{16}\bigg(\frac{\eta(4z)}{\eta(2z)}\bigg)^8+16\bigg(\frac{\eta(z)}{\eta(2z)}\bigg)^8\bigg(\frac{\eta(4z)}{\eta(2z)}\bigg)^{16}=1$$
is equal respectively, to
$$\beta_{4n}+\alpha_{4n}=1.$$
Using Ramanujan's notations for
$\eta(z)=\frac{\phi(q)}{\sqrt{2}(G_{n})^2}$
$\eta(2z)=\frac{\phi(q^2)}{\sqrt{2}(G_{4n})^2}$
$\eta(4z)=\frac{\phi(q^4)}{\sqrt{2}(G_{16n})^2}$
and that
$\frac{\phi(q)}{\phi(q^2)}=\frac{\beta_{4n}^{1/4}}{\beta_{n}^{1/8}}$
$\frac{\phi(q^2)}{\phi(q^4)}=\frac{\beta_{16n}^{1/4}}{\beta_{4n}^{1/8}}$
and moreover
$\beta_{n}=\frac{g_{n}^8}{G_{n}^8}$
$\beta_{4n}=\frac{g_{4n}^8}{G_{4n}^8}$
$\beta_{16n}=\frac{g_{16n}^8}{G_{16n}^8}$
$$4g_{4n}^8 G_{4n}^8\big(G_{4n}^8-g_{4n}^8\big)=1\tag1$$
we arrive to
$\beta_{4n}=\bigg(\frac{\eta(z)}{\eta(2z)}\bigg)^{16}\bigg(\frac{\eta(4z)}{\eta(2z)}\bigg)^8=\frac{\beta_{4n}^{5}G_{4n}^{48}}{\beta_{n}^2 G_{n}^{32} \beta_{16n}^2 G_{16n}^{16}}=$
$=\frac{g_{4n}^{40} G_{4n}^8}{g{n}^{16} G_{n}^{16} g_{16n}^{16}}=$
$=\frac{g_{4n}^{40} G_{4n}^8}{g_{n}^{16} G_{n}^{16} 2^4 g_{4n}^{16} G_{4n}^{16}}=$
$=\frac{g_{4n}^{40} G_{4n}^8}{g_{4n}^{16} g_{4n}^{16} G_{4n}^{16}}$
simplifying
$\beta_{4n}=\frac{g_{4n}^8}{G_{4n}^8}$
With analog considerations we have
$$\frac{g_{4n}^8}{ G_{4n}^8}+\frac{1}{4g_{4n}^8 G_{4n}^{16}}=1$$
or that is the same the equation (1).
