Probability: 6 Dice are rolled. Which is more likely, that you get exactly one 6, or that you get 6 different numbers? Here's the question:

6 Dice are rolled. Which is more likely, that you get exactly one 6, or that you get 6 different numbers?

Here's what I've done:

The number of possible outcomes is $6^6 = 46656$.

The probability of rolling exactly one 6 = $\frac{1}{6}\times(\frac{5}{6})^5 = \frac{3125}{46656}$

The probability of getting six different numbers is: 
$C(6,1)\times C(5,1)\times C(4,1)\times C(3,1)\times C(2,1)\times C(1,1) = \frac{720}{46656}$

Therefore if everything I've said above is true, then it is more likely that you will roll exactly one six. However I'm really not sure about the last part. Is this correct way to solve this type of problem and can the combinations part be simplified?

EDIT: Since I'm also looking for a better idea of how to solve this type of problem, rather than just this specific case, so could you please include how to solve this problem for rolling 5 dice, as well as/or instead of 6 dice in your answer, so that I can see the pattern of what is happening? Many thanks.
The question is very easy to answer without computing probabilities. Every combination with six different numbers contains exactly one six. There are then additional combinations which contain exactly one six - e.g. $111116$. So the probability of exactly one six is greater.
The trick is to compare the sets of possible outcomes for each condition. Whichever condition's result set has a higher cardinality -in other words, whichever set has more possible outcomes- is more likely to occur, and if the two sets are equal in size, then they are equally likely.

When the number of dice is the same as the number of sides per die -for example, 6d6- you can take a shortcut by showing that one condition is a proper subset of the other. This is how @MarkBennet's answer works. Because there are as many dice as sides, every outcome with all different numbers must show exactly one of each number. So as long as it's possible for your chosen number to come up at all (your example uses a 6 on a six-sided die), then every outcome with all different numbers must also have exactly one of the chose number. You can also show outcomes with exactly one six which don't have all different numbers, like (2, 2, 3, 4, 5, 6), but you can't show outcomes with all different numbers but not exactly one six. Therefore, the set of outcomes with exactly one six is larger, and therefore it's more likely to roll exactly one six.

When there are more dice than sides, you have a different shortcut: the pigeonhole principle. Every die must show some number, so since there are more dice than sides, there must always be at least one number that comes up twice. This means that the set of outcomes with all different numbers is zero, so as long as it's possible for your chosen number to come up at all, then the set of outcomes with exactly one of that number has a greater cardinality. This makes exactly one six more likely than all different numbers, because there are no possible outcomes with all different numbers.

But when there are more sides than dice, there are no shortcuts. You can still show outcomes which satisfy both conditions at the same time, like (2, 3, 4, 5, 6) on 5d6: all different numbers, and exactly one six. And you can still show outcomes with exactly one six but all different numbers: continuing our example, (3, 3, 4, 5, 6) on 5d6 is one of them. But now, you can also show outcomes containing all different numbers but not exactly one six: e.g. (1, 2, 3, 4, 5).

When there's no shortcut, you have to compare the cardinalities of both sets, which is what you did in the question (and @MattS did in his answer). Whichever set has a higher cardinality is more likely to occur.