First, let's get the chances of each of the six "__ is the highest die" probabilities of occurring.

There is 1 way to get it so no number is greater than 1; all the dice must say 1.

There is 2^n ways to get it so no number is greater than 1 because you can get 2 favorable outcomes, 2 on the die or 1 on the die. Since we want at least 1 die to say 2, we must subtract however many ways there are to get at most 1 on every dice, 1 in this case. So there are (2^n)-1 ways to get a highest number of 2.

Similar logic can be used with 2 and 3. There are 3^n ways to get no more than 3 on each die, and 2^n ways to get no more than 2 on each die, so there are (3^n)-(2^n) ways to get 3 as the highest number on any die.

This logic also applies for highest numbers of 4, 5, and 6. Since each of the n dice has 6 faces, there are 6^n different outcomes.