Expectation and variance (Branching processes)

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Suppose $N_0=1$ and define $N_n$ inductively as follows

$N_{n}=\left\{ \begin{array}{ccc} \sum_{m=1}^{N_{n-1}}\xi_{n,m} & \textrm{si} & N_{n-1}>0\\ 0 & \textrm{si} & N_{n-1}=0 \end{array}\right.,$

where the $\xi_{n,m}$ are independent and have $P(\xi_{n,m}= k)$. Here we are thinking of $N_n$ as the number of males in the $n$-th generation and we suppose that each man has an independent and identically distributed number of male children. Let $\mu=\mathbb{E}\left[\xi_{n,m}\right]$ and $\sigma^{2}=\mathbb{V}\mathrm{ar}\left[\xi_{n,m}\right]$. Find $\mathbb{E}\left[N_{n}\right]$ and $\mathbb{V}\mathrm{ar}\left[N_{n}\right]$.

MY ATTEMPT:

Using law of iterated expectations, we have:

$\mathbb{E}\left[N_{n}\right]=\mathbb{E}\left[\mathbb{E}\left(\left.N_{n}\right|N_{n-1}\right)\right]=\mathbb{E}\left[\mu N_{n-1}\right]=\mu\mathbb{E}\left[N_{n-1}\right]=\mu\mathbb{E}\left[\mathbb{E}\left(\left.N_{n-1}\right|N_{n-2}\right)\right]=\mu\mathbb{E}\left[\mu N_{n-2}\right]=\mu^{2}\mathbb{E}\left[N_{n-2}\right]=...=\mu^{n}\mathbb{E}\left[N_{0}\right]=\mu^{n}$

Using law of iterated variance, we have:

$\mathbb{V}\mathrm{ar}\left[N_{n}\right]=\mathbb{E}\left[\mathbb{V}\mathrm{ar}\left(\left.N_{n}\right|N_{n-1}\right)\right]+\mathbb{V}\mathrm{ar}\left[\mathbb{E}\left(\left.N_{n}\right|N_{n-1}\right)\right]=??$

Can you help me?