Let $\{X_{n,k} : n,k \geq 1\}$ be a collection of i.i.d. $\mathbb{Z}_+$-random variables with finite variance $\sigma^2 > 0$ and mean $\mu > 0$. Define $(Z_n)_{n\geq 0}$ recursively by $Z_0 = 1$ and
$$Z_n = \sum_{k=1}^{Z_{n-1}} X_{n,k}.$$
Prove that the process $M$ defined by $M_n = \mu^{-n}Z_n$ is an $(\mathcal{F}_n)$-martingale where $\mathcal{F}_n = \sigma(Z_0, \dots, Z_n)$.
First we show that each $M_n \in \mathcal{F}_n$. This follows trivially since $Z_n \in \mathcal{F}_n$ by construction. How do I show that $M_n \in \mathcal{L}^1$ and the martingale property
$$\mu^{-n}Z_n \overset{a.s.}{=} \mu^{-(n+1)}\mathbb{E}[Z_{n+1} | \mathcal{F}_n] ?$$
We have that
$$\mathbb{E}|M_n| = \mathbb{E}[M_n] = \mu^{-n}\sum_{k=1}^{Z_{n-1}}\mathbb{E}[X_{n,k}] = \dots ?$$
For the martingale property,
$$\mathbb{E}[M_{n+1} | \mathcal{F}_n] = \mu^{-(n+1)} \mathbb{E}[\sum_{k=1}^{Z_n} X_{n+1,k} | \mathcal{F}_n] = \dots ?$$
The main problem is that the $Z_{n}$'s in the index set of the sum confuses me, and I am not sure how to handle these. Any explanation of this problem will be appreciated.
After a bit of googling, I found that we should have $\mathbb{E}[Z_n] = \mu\mathbb{E}[Z_{n-1}] = \mu^n$ and $\mathbb{E}[Z_{n+1}|\mathcal{F}_n] = \mu Z_{n}$ but for the same reason as above I do not understand the justification.
EDIT
So,
$$\mathbb{E}[Z_{n}] = \mathbb{E}[\sum_{k=1}^{Z_{n-1}} X_{n,k}] = \sum_{j=1}^\infty\mathbb{P}(Z_{n-1} = j)\sum_{k=0}^{j}\mathbb{E}[X_{n,k}] = \mu\sum_{j=1}^\infty\mathbb{P}(Z_{n-1} = j)j = \mu\mathbb{E}[Z_{n-1}] = \mu^n\mathbb{E}[Z_0] = \mu^n.$$
Still don't get the martingale property. If we apply the same reasoning as above, we should get
$$\mathbb{E}[Z_{n+1}|\mathcal{F}_n] = \mathbb{E}[\sum_{k=1}^{Z_{n}} X_{n+1,k} | \mathcal{F}_n] = \sum_{j=1}^\infty \mathbb{P}[Z_n = j]\sum_{k=0}^j \mathbb{E}[X_{n+1, k} | \mathcal{F}_n] = \overset{?}{\cdots} = \mu Z_{n}$$
$E[Z_{n+1}]=E[X_{1,n+1}+X_{2,n+1}+\ldots+X_{Z_{n},n+1}]$
$=\sum_{k=0}^\infty P(Z_n=k)E[X_{1,n+1}+X_{2,n+1}+\ldots+X_{n+1}]$
$=\sum_0^\infty P(Z_n=k)k\mu=\mu\sum_{k=0}^\infty P(Z_n=k)k=\mu E[Z_n]$
$E[Z_{n+1}|F_n]=E[X_{1,n+1}+X_{2,n+1}+\ldots+X_{Z_{n},n+1}|F_n]$
Since $X_{1,n+1}, X_{2,n+1},\ldots$ are i.i.d. and do not depend on $F_n$ you have
$=E[X_{1,n+1}]E[Z_{n}|F_n]=\mu Z_n$, because $Z_n$ is $F_n$ measurable.