$By \ standard\ Formula\

e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}....\

=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}....\

sin (e^x-1)=sin\bigg(1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}-1\bigg)\

=sin\bigg(x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}.....\bigg)\

But sinx'=x'-\dfrac{x'^3}{3!}+\dfrac{x'^5}{5!}.........\

\therefore\sin\bigg(x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}....\bigg)\

$ $=\bigg(x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}..\bigg)- \dfrac 16\bigg(x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}.\bigg )^3........\

=x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}.. -\dfrac 16\bigg(x^3-3\dfrac{x^4}2\bigg)\

=x+\dfrac{x^2}2+\dfrac{x^3}6-\dfrac{x^3}6- \dfrac 5{24}x^4\

=x+\dfrac{x^2}2- \dfrac{ 5x^4}{24}-------Proved $