Hint: Note that the sum on the left is a geometric series. Without worrying yet about division by $0$, we get that the sum is $\frac{1-e^{2^nit}}{1-e^{it}}.$ For the induction, perhaps use the fact that $1-e^{2^{n+1}it}=(1-e^{2^{n}it})(1+e^{2^{n}it}).$
Once you have worked through the idea, note the following slicker but equivalent calculation. There are two cases, $e^{it}=1$ and $e^{it}\ne 1$. In the second case, multiply each side by $1-e^{it}$. Admire the beautiful collapse on both the left-hand side and the right-hand side.
For the right-hand side, which is less familiar, look for example at $(1+a)(1+a^2)(1+a^4)(1+a^8),$ and multiply on the left by $1-a$.