Power series raised to an exponent...where does Wikipedia get this formula? On Wikipedia, they claim that
$$
\left(\sum_{k=0}^\infty a_k x^k\right)^N
$$
is another power series with $c_0 = a_0^N$ and $$c_m = \frac{1}{ma_0} \sum_{k=1}^m (kN-m+k) a_k c_{m-k}.$$  I tried proving this by induction but it's an absolute nightmare.  Can someone provide some kind of motivation for where this power series expansion comes from?
 A: That formula follows from the form of the derivative:
$$
c'(x)=N·a(x)^{N-1}·a'(x)\implies a(x)·c'(x)=N·c(x)·a'(x)
$$
and then looking at the coefficients of $x^{m-1}$
$$
\sum_{j=1}^m a_{m-j}·(j)c_{j}=N·\sum_{j=0}^{m-1}c_j(m-j)a_{m-j}
\implies
ma_0·c_m=\sum_{j=0}^{m-1} (Nm-Nj-j)a_{m-j}c_j
$$
or set $k=m-j$ to obtain the given formula.
A: If you want a non-recursive solution you may be interested in this. If we let $f(x) = \sum_{k=0}^{\infty}a_k x^k$, then we have that
\begin{align*}
f(x)^n &= \left( \sum_{k=0}^{\infty}a_k x^k \right)^n \\
       &= \sum_{k=0}^{\infty} \left(\sum_{\substack{0 \leq r_1,\ldots,r_n \leq k \\ r_1 + \cdots + r_n = k}}a_{r_1} \cdots a_{r_n}\right) x^k \\
\end{align*}
Now for any specific choice of $r_1,\ldots,r_n$ satisfying the condition
$$0 \leq r_1,\ldots, r_n \leq k,\quad r_1 + \cdots + r_n = k$$
there are, say, $m \leq n$ many distinct elements in the set $S = \{r_1,\cdots,r_n\}$, call these distinct elements
$$s_1,\ldots,s_m$$
and define
$$N(s_i) = \# \text{ of times $s_i$ appears in $(r_1,\ldots,r_n)$}$$
then we see that the number of times
$$a_{r_1}\cdots a_{r_n}$$
is counted is equal to
$$C(r_1,\ldots,r_n) := \frac{n!}{N(s_1)! \cdots N(s_m)!} \quad \text{(The multinomial coefficient)}.$$
Thus we have
$$f(x)^n = \sum_{k = 0}^{\infty} \left( \sum_{\substack{0 \leq r_1 \leq\ldots \leq r_n \leq k \\ r_1 + \cdots + r_n = k}}C(r_1,\ldots,r_n)a_{r_1} \cdots a_{r_n} \right) x^k$$
Notice that now the inner sum is over a smaller set and hence (in theory) makes the calculation easier.
