What is the probability that if we roll one six-sided die for n times - whatever the number we see in the first roll, no other following roll has a number higher than that (that is all the numbers we see in the following rolls are less than or equal to the first number)?
Example:
We can consider some sample cases to better explain the question. We roll ONE 6-sided die 5 times
1 1 1 1 1 -> Success
5 4 3 2 1 -> Success
4 4 3 2 4 -> Success
5 4 3 2 6 -> Fail
If the first number is $1$, then the probability is $\left(\frac{1}{6}\right)^{n-1}$
If the first number is $2$, then the probability is $\left(\frac{2}{6}\right)^{n-1}$
If the first number is $3$, then the probability is $\left(\frac{3}{6}\right)^{n-1}$
And so on.
So the required probability is $\sum_{k=1}^6 \frac{1}{6}\left(\frac{k}{6}\right)^{n-1}$, or more simply $\frac{1}{6^n}\left(1^{n-1}+2^{n-1}+\cdots+6^{n-1} \right)$.