$(e^{ix}-1)^{2n}-(e^{-ix}-1)^{2n}=-(e^{inx}+e^{-inx})(e^{ix}-2+e^{-ix})^n$.
I attempted to write those awful $e$'s into trigonomitric functions, but I failed to prove it. Wolframalpha did not interpret it properly.
Edit Thanks for helping me. The correct result should be $$(e^{ix}-1)^{2n}-(e^{-ix}-1)^{2n}= ( e^{2ix} -2e^{ix}+1)^{n}-(e^{-2ix} -2e^{-ix}+1)^{n}\\= ( e^{2ix} -2e^{ix}+1)^{n}-\frac{(e^{2ix} -2e^{ix}+1)^{n}}{e^{2inx} }=( e^{2ix} -2e^{ix}+1)^{n}\cdot(1-e^{-2inx})=e^{inx}(e^{ix} -2+e^{-ix})^{n}\cdot(1-e^{-2nix})=(e^{ix} -2+e^{-ix})^{n}\cdot(e^{inx}-e^{-inx})$$
I obtain a different result
$$(e^{ix}-1)^{2n}-(e^{-ix}-1)^{2n}= ( e^{2ix} -2e^{ix}+1)^{n}-(e^{-2ix} -2e^{-ix}+1)^{n}\\= ( e^{2ix} -2e^{ix}+1)^{n}-\frac{(e^{2ix} -2e^{ix}+1)^{n}}{e^{2inx} }=( e^{2ix} -2e^{ix}+1)^{n}\cdot(1-e^{-2inx})=e^{inx}(e^{ix} -2+e^{-ix})^{n}\cdot(1-e^{-2nix})=(e^{ix} -2+e^{-ix})^{n}\cdot(e^{inx}-e^{-inx})$$