Let $\mu$ and $\sigma^2$ be the mean and variance of the number of offspring of one individual.
$\mu=\Sigma_{j=0}^{\infty}jP_j$
We want $E[X_n]$
The note is as follow:
Note that $X_n = \Sigma_{i=1}^{n-1}Z_i$,where $Z_i$ is the number of offspring of the $i$th individual.
$E(X_n) = E(E(X_n|X_{n-1})) = E(E(\Sigma_{i=1}^{n-1}Z_i|X_{n-1})) = $
$E(X_{n-1}\mu|X_{n-1}) = E(X_{n-1}\mu) = \mu$$E(X_{n-1})$
Since$X_0=1$,we get $E(X_n) = \mu^n,n=1,2,...$
How do we get from $E(E(\Sigma_{i=1}^{n-1}Z_i|X_{n-1}))$equal to$E(X_{n-1}\mu|X_{n-1}) $
Your summation upper index should be $X_{n-1}$, not $n-1$. Think about it; at time $n-1$, I have $X_{n-1}$ individuals, so each of them are going to have a $Z_i$ offsprings. Thus, if you condition on the number of individuals of the previous step, you have $X_{n-1}$ iid $Z_i$s.