Simplest way to see that the Euler density is topological

Question

I’m looking for a simple way to see that the integrated Euler density $\sqrt{g} \, E_{2n}$ is topological (i.e. metric-independent in general even dimension $2n$). I can see it it must be true in two dimensions since $E_2 = R$ and because of the two-dimensional relation between the Ricci and scalar curvature $R_{\mu\nu} = (R/2)g_{\mu\nu}$ which implies that the variation vanishes \begin{equation} \delta \int \mathrm{d}^{2}x \sqrt{g} \, E_2 = \int \mathrm{d}^{2}x \sqrt{g} \left(R_{\mu\nu} - \frac{1}{2}R g_{\mu\nu}\right) \delta g^{\mu\nu} = 0 \end{equation} I suspect that the key to general dimension lies in the formula \begin{equation} E_{2n} = \frac{1}{2^n} {R^{i_1j_i}}_{k_1l_1} \cdots {R^{i_nj_n}}_{k_nl_n} \epsilon_{i_1 j_1 \cdots i_n j_n} \epsilon^{k_1l_1\cdots k_n l_n} \end{equation} but it’s unclear how to proceed other than brute force.