Problem
Suppose we roll 6 dice. We want them to not be all the same number, we don't care what number or how many dice are different. What is the probability of this event?
Attempt
We are simply looking for the probability that all the dice are not the same number. So if take the complement of the probability that they are all the same we get
$$1- \bigg(\frac{1}{6}\bigg)^6$$
Is this correct or am I leaving out a factor of $6$?
The total number of possible combinations of numbers we can get from rolling $6$ fair dice is $6^6$. Out of all these possibilities, there are only $6$ outcomes where all the dice display the same number (these are the cases $(1,1,1,1,1,1); (2,2,2,2,2,2)$, etc.). Therefore, there are $6^6-6$ possibilities where the numbers are not all the same.
Finally, we find the probability of all the numbers being different by calculating the ratio between the total number of desired outcomes with the total number of possible outcomes.
This is: $\frac{6^6-6}{6^6}=1-\frac{6}{6^6}=1-\frac{1}{6^5}$
Therefore, if we compare this with your attempted solution, we see that your attempt is not correct (as you correctly identify), since you are out by a factor of $6$.