Let $N_n$ be the number of throws before all $n$ dice have shown $6$. Set $m_n := E[N_n]$. Write a recursive formula for $m_n$. Suppose we throw $n$ independent dices. After each throw we put aside the dices
showing $6$ and then perform a new throw with the dices not showing $6$, repeating the process until all dices have shown $6$. Let $N_n$ be the number of throws before all dice have shown $6$. Set $m_n := E[N_n]$ and let $Y$ denote the number of dices not showing $6$ after the first throw.
I've showed that $Y \sim b(n,\frac 5 6)$, because we can consider the throw as $n$ independent trials with succes probability $\frac 5 6$.
Now I must write a recursive formula for $m_n$ and decide $m_1, \ldots, m_5$.
Using the definition of expectation of a discrete RV: 
$E[N_n] = 1 \cdot P(N_n = 1) + 2 \cdot P(N_n = 2) + \ldots$
But this is not very recursive - Any suggestions ?

Consider the first throw. Given that we get no six, the expectation is $1+m_{n}$. Given that we get exactly $1$ six, the expectation is $1+m_{n-1}$. Given that we get $2$ sixes, the expectation is $1+m_{n-2}$, and so on. Thus
$$m_n=\sum_{k=0}^n (1+m_{n-k})\binom{n}{k}\left(\frac{1}{6}\right)^{k}\left(\frac{5}{6}\right)^{n-k} .$$
We can simplify this a bit, by solving for $m_n$, and combining the "$1$" terms. There may be further simplification. 
Remark: We assumed that for example the first throw of the $n$ dice counts as $1$ throw. If throwing $k$ dice counts as $k$ throws, then the problem is simpler.
