Series expansion of $e^{sinx}$ and set partition In this thread, Michael Hardy derived a general formula for $e^{f(x)}$. He talked about set partition:
Need help to solve taylor series of $e^{\sin x}$

I just wonder how I can count the set partitions and continue to write in more terms.
I wish to know how to count the set partitions. How do you count it along the integer partition. This is important because you can write the coefficient for the taylor expansions of $e^{f(x)}$
 A: A partition of a set $S$ is a set of disjoint subsets of $S$ whose union is $S$.  To put it another way, a partition is the set of equivalence classes of some equivalence relation on $S$.  
It seems to me that there are a couple of typos in the image you copied.  (By the way, please avoid posting images when possible.  Images can't be browsed and may not be accessible to those using screen readers.  It would be much better to copy and paste the parts of the post you wanted, I think.)
Any way, let's look at the example, the number of ways to partition a set with four elements.  I will use the terminology $d$-set to mean a set with $d$ elements.  
There is only one way to partition it into a $4$-set: take all the elements.  
There are $4$ ways to partition it into a $1$-set and a $3$-set, because there are $4$ ways to choose the singleton set, and then the $3$-set is determined.  
There are $3$ ways to partition it into two $2$-sets.  There are ${4\choose2}=6$ ways to choose two elements, but if we choose $\{a,b\}$ first, we are left with $\{c,d\}$ as the second set, just as if we had chosen $\{c,d\}$ first, so we divide by $2$ in order to avoid double counting.
There are $6$ ways to partition it into a $2$-set and two $1$-sets.  There are ${4\choose2}=6$ ways to choose the $2$-set, and once we do that, the two singleton sets are determined.
Finally, there is only $1$ way to partition it into $4$ singleton sets.
As to typos, in the formula for $e^{f(x)}$ the $x^2$ in the quadratic term is missing.  In the $x^4$ term, the summand should be $6a_2a_1^2$ instead of $6a_2a_1$.  I haven't checked the other terms, but I wouldn't accept them at face value.
If you are just interested in calculating a few terms of the power series, another approach may be easier.
$$\begin{align}
e^{\sin x}&=\exp\left(\sum_{n=0}^\infty (-1)^{n} x^{2n+1}/(2n+1)!\right)\\
          &=e^xe^{-x^3/3!}e^{x^5/5!}\cdots\\
          &=\left(1+x+{x^2\over2!}+\cdots\right)
\left(1-{x^3\over3!}+{1\over2!}\left({x^3\over3!}\right)^2+\cdots\right)
\left(1+{x^5\over5!}+{1\over2!}\left({x^5\over5!}\right)^2+\cdots\right)
\end{align}$$ 
Now it isn't hard to pick out the coefficients of $x^n$, so long as $n$ isn't too big.  Obviously, the constant term is $1$.  The linear and quadratic terms can only come from the first factor, so we have $x+{x^2\over2!}$.  There is no contribution to the $x^3$ or $x^4$ terms from factors beyond the second, so we get $$\left(\frac{1}{3!}-\frac{1}{3!}\right)x^3+\left(\frac{1}{4!}-\frac{1}{3!}\right)x^4$$ 
and so on.
