I am trying to prove the following basic fact about exponentiation for a version of binomial coefficients for formal power series, based on R.P. Stanley's definition in "Enumerative Combinatorics Volume 1," Example 1.1.10. The book simply says "all the expected properties of exponentiation are indeed valid," so I am assuming it's supposed to be obvious, but I have not been able to prove it.
Let $F(x) \in \mathbb{C}[[x]]$ be a formal power series such that $F(0) = 0$ and $\lambda \in \mathbb{C}$. Define
$(1 + F(x))^\lambda := \sum_{n\ge0}$ ${\lambda}\choose{n}$$F(x)^n$, where ${\lambda}\choose{n}$$:= \lambda(\lambda-1)\cdots (\lambda-n+1)/n!$.
The problem: Prove for all $\lambda,\mu \in \mathbb{C}$, $((1 + F(x))^\lambda)^{\mu}$ = $(1+F(x))^{\lambda\mu}$.
My attempt thus far:
I tried to expand based on the definition above and obtained
$((1+F(x)^\lambda)^\mu = \sum_{n \ge 0}$$\mu \choose n$$(\sum_{i\ge1}$$\lambda\choose i$$F(x)^i)^n$.
I want to show that this is equal to
$\sum_{n\ge0}$$\lambda\mu\choose n$$F(x)^n$.
Looking at individual coefficients for each $F(x)^n$, I "reduced" the problem to showing that
$\mu\choose1$$\lambda\choose n$ $+$ $\mu\choose 2$ [ $\lambda \choose 1$$\lambda \choose n-1$ $+$ $\lambda \choose 2$$\lambda \choose n-2$ $+ \dots +$ $\lambda \choose n-1$$\lambda \choose 1$ ] $+ \dots +$ $\mu \choose n$ $\underset{n}{\underbrace{{\lambda \choose 1} \cdots {\lambda \choose 1}}}$ $\;\; = \;\; $ $ \mu\lambda \choose n$.
However, I don't know if this the right approach! There must be a simpler way! Any insight is appreciated.
Prove that $\, ((1+x)^\lambda)^\mu = (1+x)^{\lambda\mu} \,$ using $\, (1+x)^\lambda = \exp(\log(1+x))^\lambda = \exp(\lambda\log(1+x)). \,$ Then substitute $\, F(x) \,$ for $\, x.\,$