Roll a die $n$ times ($n$ is a natural number). What is the probability that 1 and 6 are observed at least once?

Note: This is a homework problem so I cannot accept solutions. I would like suggestions as to how to proceed.

I have that each trial of rolling a die is independent. So I can say:

Let $P(1)$ = Rolling a 1 and $P(6)$ = Rolling a 6. We want to find $P(1\cap 6)$. Since rolling a die and getting an outcome in a trial is independent of other trials: $P(1\cap6)=P(1)P(6) $

It's unclear to me how to find the probability that you roll a 1 if you roll $n$ times. Would it be $P(1)=1/6^n$

Answer

The total number of combinations is $6^n$.

The number of combinations which include only $[2,3,4,5,6]$ is $5^n$.

The number of combinations which include only $[1,2,3,4,5]$ is $5^n$.

The number of combinations which include only $[2,3,4,5]$ is $4^n$.

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So the probability of not observing $1$ and $6$ is therefore:

$$\frac{5^n+5^n-4^n}{6^n}$$

And the probability of observing $1$ and $6$ at least once is:

$$1-\frac{5^n+5^n-4^n}{6^n}$$