So you want to know the probability of getting all the faces at least once after rolling the die $n$ times. It is convenient to introduce the number $N_k$ of faces that have been seen after $k$ steps. Obviously, we have $N_1=1$. Also, $N_{k+1}=N_k$ with probability $\frac{N_k}{6}$ and $N_{k+1}=N_k+1$ otherwise -- in other words, the process $\{ N_k \}_{k \geq 1}$ is an Markov chain. One can thus easily compute the vector $V_k=(\mathbb{P}[N_k=1],\mathbb{P}[N_k=2], \ldots, \mathbb{P}[N_k=6])$ for $k=1,2, \ldots$ and solve the problem. One finds $V_{n+1} = V_0 \, A^{n}$ where $V_0=(1,0,\ldots,0)$ and $A$ is the transition matrix of the Markov chain:
$$A = \begin{pmatrix}
1/6 &5/6 &0 &0 &0 &0 \\
0 &2/6 &4/6 &0 &0 &0\\
0 &0 &3/6 &3/6 &0 &0 \\
0 &0 &0 &4/6 &2/6 &0 \\
0 &0 &0 &0 &5/6 &1/6\\
0 &0 &0 &0 &0 &1
\end{pmatrix}$$
To find $V_n$, diagonalize $A$ and then compute the powers. This gives
$$V_{n+1} = \frac{1}{6^{n+1}}\begin{pmatrix}1\\-5\\10\\-10\\5\\1\end{pmatrix}^{tr}
\begin{pmatrix}
6^n &0 &0 &0 &0 &0 \\
0 &5^n &0 &0 &0 &0 \\
0 &0 &4^n &0 &0 &0 \\
0 &0 &0 &3^n &0 &0 \\
0 &0 &0 &0 &2^n &0\\
0 &0 &0 &0 &0 &1
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & -1 & 1 \\
0 & 0 & 0 & 1 & -2 & 1 \\
0 & 0 & -1 & 3 & -3 & 1 \\
0 & 1 & -4 & 6 & -4 & 1 \\
-1 & 5 & -10 & 10 & -5 & 1
\end{pmatrix}
$$
For example, after rolling a die 7 times, set $n=6$ in the preceding formula to get
$$V_7 = \begin{pmatrix}6,1890,36120,126000,100800,15120\end{pmatrix} / 6^7$$
From left to right, these are the chances of having observed exactly 1, 2, ..., through 6 faces. The chance of having seen all 6 faces is the last entry, $15120/6^7 = 35/648 \approx 0.054$. In general, the last entry of $V_{n+1}$ equals
$$\Pr[\text{All faces seen after } n+1 \text{ throws}] = 1-5\ 2^{2-n}+5\ 3^{1-n}(1+2^n)-6^{1-n}(1+5^n).$$
