Power series of $f(z)^8$ Goal: If $f(z)^8$ is analytic on some domain D and if $f(z)$ is continuous on domain $D$ with $f(0) = 0$, then the power series $f(z)^8 = \sum a_nz^n$ will begin with $n$ divisible by $8$ 
My attempt: 
Allow $g(z) = f(z)^8$ which is analytic on some domain D. Then, we use Cauchy integral to calculate our $a_n$.
$$g^{(n)}(z) = a_n = \frac{n!}{2\pi i}\int_{D} \frac{g(w)}{(w-z)^{n+1}} dw$$
Now, I notice that we can split the integrand up, but I'm not quite sure if it's helpful
$$\frac{n!}{2\pi i} \int_D \frac{1}{(w-z)^{j}} \frac{f(w)^8}{(w-z)^{8k+1}} dw$$. 
Where $j + 8k + 1 = n + 1$
Can anyone tell me if this is going in the right direction or any hints?
 A: Your assumption of continuity of $f$ along with the analyticity of $f^8$ is enough to conclude $f$ is analytic. For example, see the following:


*

*If $z\mapsto f(z)^n$ is analytic then $f$ is analytic

*If $f^2$ is an analytic function then so is $f$
The former gives the result immediately but is a tad tougher to prove, while the latter gives the result after noting $f$ being continuous implies $f^2$ and  $f^4$ are continuous, at which point we apply the latter result three times.
We thus know $f$ is analytic, so $$f(z) = \sum_{k=0}^\infty c_kz^k$$
Applying your edit that $f(0)=0,$ we must have $c_0 = 0.$ Let us first assume $c_1 \neq 0,$ so $$f(z) = \sum_{k=1}^\infty c_kz^k = c_1 z + \mathcal{O}(z^2)$$
Thus, by the Binomial Theorem, we have that
$$f(z)^8 = (c_1 z + \mathcal{O}(z^2))^8 = (c_1 z)^8 + \mathcal{O}(z^9)$$
Which proves the claim whenever $c_1 \neq 0.$
We now use this idea to prove the general case. Suppose $c_k = 0$ for $0 \le k \le n-1,$ but $c_n \neq 0.$ Then:
$$f(z)^8 = \left(\sum_{k=n}^\infty c_kz^k\right)^8 = (c_n z^n + \mathcal{O}(z^{n+1}))^8 = (c_n z^n)^8 + \mathcal{O}(z^{8(n+1)})$$
Which proves that the leading nonzero term will have a power divisible by $8.$ This of course extends readily to other powers of $f.$
