Question: 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 ?

Answer: 
$$m\_{n}=\mathbb{E}N\_{n}=\sum\_{y=0}^{n}\mathbb{E}\left(N\_{n}\mid Y=y\right)P\left(Y=y\right)$$
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: $$m\_{n}=\frac{6^{n}+\sum\_{y=1}^{n-1}m\_{y}\binom{n}{y}5^{y}}{6^{n}-5^{n}}$$

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**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: $$\mathbb{E}N\_{n}=\sum\_{k=0}^{\infty}P\left(N\_{n}>k\right)=\sum\_{k=0}^{\infty}\left[1-\left(1-\left(\frac{5}{6}\right)^{k}\right)^{n}\right]$$
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.