Answer
$1+x+\dfrac{x^2}{2}-\dfrac{x^4}{8}+...$
Work Step by Step
We have the Maclaurin series for $e^x$ as follows:
$$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$$
Replace $\sin x$ for $x$ in the above form.
Therefore, we have:
$f(x)=e^{\sin x}=1+\sin x+\dfrac{\sin^2 x}{2!}+\dfrac{\sin^3 x}{3!}+\dfrac{\sin^4 x}{4!}+...\\=1+x-\dfrac{x^3}{6}+\dfrac{1}{2}(x^2-\dfrac{x^4}{3})+\dfrac{1}{6} x^3+...\\=1+x+\dfrac{x^2}{2}-\dfrac{x^4}{8}+...$
