Rademacher functions form an orthonormal system but not an orthonormal basis

I would like to know how to show that the functions $$r_n(t)=\operatorname{sgn}\big(\sin(2^n \pi t)\big)$$ (where $\operatorname{sgn}$ is the sign function) form an orthonormal system but not an orthonormal basis from $L_2([0,1])$.

Progress: I pick two different variables $i,k$ and show that $\langle r_i (t),r_k (t)\rangle =0$, so I can take the integral $\int_0^1 r_i (t) r_k (t)\,dt$, but I do not know how to show that this is 0.

To compute $\langle r_n, r_m\rangle$ with $n<m$, say, observe that $$\langle r_n, r_m\rangle=\int_0^1 \operatorname{sgn}(\sin(2^n\pi t))\operatorname{sgn}(\sin(2^m\pi t))\,dt\\=\sum_{k=0}^{2^n-1}\int_{k2^{-n}}^{(k+1)2^{-n}} \operatorname{sgn}(\sin(2^n\pi t))\operatorname{sgn}(\sin(2^m\pi t))\,dt$$ In each summand, $\operatorname{sgn}(\sin(2^n\pi t))$ is constant and $\operatorname{sgn}(\sin(2^m\pi t))$ runs over $2^{m-n}$ full periods. Thus each summand is zero.

Remark: What is still left for you to show that this orthnormal system fails to be a basis?