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 ?
 A: Two closed formulas, not based on the recursion you suggest:
$$
E(N_n)=\sum_{k\geqslant0}\left(1-\left(1-\left(\frac56\right)^k\right)^n\right)
$$
$$
E(N_n)=\sum_{i=1}^n{n\choose i}\frac{(-1)^{i+1}}{1-\left(\frac56\right)^i}
$$
A: 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.
A: $$m_{n}=\mathbb{E}N_{n}=\sum_{y=0}^{n}\mathbb{E}\left(N_{n}\mid Y=y\right)P\left(Y=y\right)$$
Here $\mathbb{E}\left(N_{n}\mid Y=y\right)=m_{y}+1$, $m_{0}=0$ and
applying the distribution of $Y$ this gives: $$m_{n}=\sum_{y=0}^{n}\left(m_{y}+1\right)\binom{n}{y}\left(\frac{5}{6}\right)^{y}\left(\frac{1}{6}\right)^{n-y}$$
and leads to:$$m_{n}=\sum_{y=0}^{n}m_{y}\binom{n}{y}\left(\frac{5}{6}\right)^{y}\left(\frac{1}{6}\right)^{n-y}+1=\sum_{y=1}^{n-1}m_{y}\binom{n}{y}\left(\frac{5}{6}\right)^{y}\left(\frac{1}{6}\right)^{n-y}+m_{n}\left(\frac{5}{6}\right)^{n}+1$$
resulting in: $$m_{n}=\frac{6^{n}+\sum_{y=1}^{n-1}m_{y}\binom{n}{y}5^{y}}{6^{n}-5^{n}}$$

addendum (in order to say something about the distribution of $N_n$)
Give the dice the numbers $1,2,\dots,n$ and let $D_{i}$ denote the
number of throws that are needed for the die with number $i$ to produce
a $6$. 
Note that $N_{n}\leq k$ if and only if $D_{i}\leq k$ for
$i=1,\dots,n$. That means that: $$P\left(N_{n}\leq k\right)=P\left(D_{1}\leq k\right)\times\cdots\times P\left(D_{n}\leq k\right)$$
Here $P\left(D_{i}\leq k\right)=1-P\left(D_{i}>k\right)=1-\left(\frac{5}{6}\right)^{k}$
for every $i$  leading to: $$P\left(N_{n}\leq k\right)=\left(1-\left(\frac{5}{6}\right)^{k}\right)^{n}$$
In one of my comments I stated that it would be 'quite a job to find
the distribution of $N_{n}$' but I was wrong there. In fact Did  made
use of this on a very elegant way in his answer. In general if $X$ is an rv taking values on positive
integers then: $$\mathbb{E}X=\sum_{k=0}^{\infty}P\left(X>k\right)$$
Applying that here leads to: $$\mathbb{E}N_{n}=\sum_{k=0}^{\infty}P\left(N_{n}>k\right)=\sum_{k=0}^{\infty}\left[1-\left(1-\left(\frac{5}{6}\right)^{k}\right)^{n}\right]$$
