<s>[INST] Taylor expansion of $\exp(\sin x)$ around $c=0$ 
Consider the Taylor's expansion around $c=0$
  and find the first 4 terms for the function $\exp(\sin x)$.

I have done this but I'm not sure if is correct.
$$e^{\sin x}=\left(1+x+\frac{x^2}{2!}+\dots+\frac{x^{n-1}}{(n-1)!}\right)^{x-\frac{x^3}{3!}+\frac{x^5}{5!}+\dots+(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!}}$$
 [/INST] You want terms up to $x^3$. The exponential is $e^t=1+t+\frac{t^2}2+\cdots$ and the sine $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$. Hence you only need to expand from two terms of the sine, as the others already exceed the degree $3$.
Now, only expanding the necessary terms
$$\begin{align}1&\to1\\
t&\to x-\frac{x^3}6\\
\frac{t^2}2&\to\frac{x^2}2-\cdots\\
\frac{t^3}6&\to\frac{x^3}6-\cdots
\end{align}$$
and in total
$$1+x+\frac{x^2}2+0\,x^3.$$

With a little more courage, up to degree $5$:
$$\begin{align}1&\to1\\
t&\to x-\frac{x^3}6+\frac{x^5}{120}\\
\frac{t^2}2&\to\frac{x^2}2-\frac{x^4}6\cdots\\
\frac{t^3}{6}&\to\frac{x^3}{6}-\frac{x^5}{12}\cdots\\
\frac{t^4}{24}&\to\frac{x^4}{24}\cdots\\
\frac{t^5}{120}&\to\frac{x^5}{120}\cdots
\end{align}$$
gives
$$1+x+\frac{x^2}2-\frac{x^4}8-\frac{x^5}{15}.$$
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