Hint: Calculate how many monotonic sequences there are that "use" only $k$ numbers.
How many surjective and non-decreasing sequences of length $n$ of elements in $\{1, \ldots, k\}$ are there?
If $(a_1, \ldots, a_n)$ is such a sequence, then $a_1=1$, $a_n=k$ and there must be exactly $k-1$ different $i$, $1\le i\le n-1$ with $a_{i+1}=a_i+1$.
Thus the number of such sequences is $n-1\choose k-1$.
How many non-decreasing sequences of length $n$ of elements in $\{1, \ldots, 6\}$ with exactly $k$ different numbers are there?
We can choose $k$ out of 6 numbers and then have $n-1\choose k-1$ sequences as in the previous paragraph. Hence the number of such sequences is ${6\choose k}\cdot {n-1\choose k-1}$.
The conditional probability of throwing a sequence containing all 6 numbers, given that the sequence is nondecresing is therefore
$$p=\frac{n-1\choose 5}{{6\choose 1}{n-1\choose 0}+{6\choose 2}{n-1\choose 1}+{6\choose 3}{n-1\choose 2}+{6\choose 4}{n-1\choose 3}+{6\choose 5}{n-1\choose 4}+{6\choose 6}{n-1\choose 5}} \\
=\frac{\frac{1}{120} n^5
- \frac{1}{8} n^4
+ \frac{17}{24} n^3
- \frac{15}{8} n^2
+ \frac{137}{60} n
- 1
}{
\frac{1}{120} n^5
+ \frac{1}{8} n^4
+ \frac{17}{24} n^3
+ \frac{15}{8} n^2
+ \frac{137}{60} n
+ 1
}\\
=1-30\cdot\frac{n^4
+ 15 n^2
+ 8
}{n^5
+ 15 n^4
+ 85 n^3
+ 225 n^2
+ 274 n
+ 120
}
,$$
which converges $\to1$ as $n\to\infty$.