TITLE: Clever derivation of $\arcsin(x)$ Taylor series
QUESTION [13 upvotes]: I was working the other day in the Math Help Centre, trying to help some first years with a calculus problem.  The problem involved investigating the Taylor series of $\arcsin(x)$.  Once the students had derived
$$\arcsin(x)=\sum_{n=0}^\infty \frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1}$$
they were asked to rederive it in a different way:

Determine the sequence $\{c_n\}_{n\in\mathbb{N}}$ such that $$x=\sum_{n=0}^\infty c_n \left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)^n.$$

Here's what I got: I recognized that the object in parentheses is the Taylor series of $\sin(x)$, so the idea is to let $\arcsin(x)=\sum_{n=0}^\infty c_nx^n$ and note that $\arcsin(\sin(x))=x$.  After that it's just an issue of computing the $c_n$'s.
There's an obvious brute-force way to do it, where for each $n$ you say "indices larger than $n$ don't matter, so now it's a finite problem."  Expand the relevant terms to get relations involving the $c_i$'s that you've already worked out.  The problem is that this will only work for finitely many values, and it was hard to determine a pattern.
Is there and obvious pattern I'm missing?  It occurred to me that the "relevant terms" that contribute to $c_n$ depend on the divisors on $n$, is this intuition correct?  
Most importantly, what is the best way to solve the problem of computing $c_n$?

REPLY [8 votes]: Here's a concise argument, but it uses $\sin^2 x+\cos^2x=1$, which isn't exactly obvious from the power series. Differentiating
$$x=\sum_{n=0}^\infty c_n\sin^nx$$
twice yields 
$$
\begin{eqnarray}
0
&=&
\sum_{n=0}^\infty c_n\left(n(n-1)\sin^{n-2}x\cos^2x-n\sin^nx\right)
\\
&=&
\sum_{n=0}^\infty c_n\left(n(n-1)\sin^{n-2}x-n^2\sin^nx\right)\;,
\end{eqnarray}
$$
which gives the recurrence
$$
\begin{eqnarray}
c_{n+2}
&=&
\frac{n^2}{(n+1)(n+2)}c_n
\\
&=&
\frac{n}{n+2}\frac{n}{n+1}c_n\;.
\end{eqnarray}$$
The factors of $n$ and $n+2$ cancel except for the final $n+2$, and with $c_0=0$ and $c_1=1$ this leads to
$$c_{2n+1}=\frac{(2n)!}{(2^nn!)^2(2n+1)}\;.$$
I got the idea for this here (item 4.3, third proof).