The Maclaurin series for $e^{\sin(x)}$ up to the fourth term I am trying to get the Maclaurin series for $e^{\sin(x)}$ up to the fourth term, which should be:
$$e^{\sin(x)}=1+x+\frac{1}{2}x^2-\frac{1}{8}x^4+o(x^4)\tag{1}$$
This is how I go about it:
$$e^{\sin(x)}=1+\sin(x)+\frac{1}{2}(\sin(x))^2+\frac{1}{6}(\sin(x))^3+o(x^4)\tag{2}$$
$$\sin(x)=x-\frac{x^3}{6}+o(x^4)\tag{3}$$
$$(\sin(x))^2=x^2-\frac{x^4}{6}+o(x^5)-\frac{x^4}{6}=x^2-\frac{x^4}{3}+o(x^5)\tag{4}$$
$$(\sin(x))^3=x^3+o(x^4)\tag{5}$$
Then I plug (3),(4) and (5) into (2), which results in:
$$e^{\sin(x)}=1+\left(x-\frac{x^3}{6}\right)+\frac{1}{2}\left(x^2-\frac{x^4}{3}\right)+\frac{1}{6}\left(x^3\right)+o(x^4)=1+x+\frac{1}{2}x^2-\frac{1}{6}x^4+o(x^4)\tag{6}$$
Which is wrong. Please tell me where I go wrong.

In my humble opinion, you could do it faster
$$A=e^{\sin(x)}\implies \log(A)=\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$
$$A=e^{\log(A)}=1+x+\frac{x^2}{2}-\frac{x^4}{8}+O\left(x^5\right
   )$$
