The different outcomes do not have the same probability of $\frac1{252}$. For example "five sixes" has probability $\frac1{6^5}=\frac 1{7776}$.
You can increase your chance of winning by predicting one of the more common outcomes.
For example "1, 2, 3, 4, 5" has a probability of $\frac{24}{3125}$, about twice as much as $\frac1{252}$.
And by the way, your calculation at best calculates the probability that someone first predicts one special outcome and then this outcome happens; what you want is that the prediction someone makes comes true.
I won't give you an actual answer, and this post is not a hint to solve the given question, but I will suggest a way that you can simplify the problem in order to understand better what's being asked.
(By the way, that's often a very useful approach, that is, simplifying the problem to get a better understanding of it)
Say you have a square and you assign numbers to the edges, in a clockwise manner, starting from the top edge. There are 4! = 24 different assignments you could make. Here are a few examples: 1234, 3241, 1324, 4123, 4321, and so on.
Now, note that the assignments 1234, 2341, 3412, and 4123 don't actually represent different squares because you could rotate them to match one another. You could put them into a grouping of their own. On the other hand, 1234 and 1324 are different assignments in the sense that no matter how you rotate either of the resulting squares, you won't be able to make them match. So, 1234 and 1324 must belong to different groupings.
What the question about the dice is asking is how many such groupings there are, but with a cube and its faces, instead of a square and its edges. Note that you can rotate a cube in more ways than you can rotate a square.
Edit: Based on the comment exchange below, here's the brute-force solution for the analogous problem with triangles. It's brute-force because I'm actually listing all the possible assignments and explicitly testing if they're pairwise equivalent or not. You will not want to do that in the dice problem. That's not how you want to solve that problem. I'm only brute-forcing the solution here to make it easier for you to understand what the problem is asking.
You can divide the 3! = 6 possible assignments into only 2 groups where every assignment in one group is equivalent to the other assignments in the same group but different from all assignments in the other group. The solution to "how many such groups are there?" is, then, 2. One is coloured blue, the other red in the figure below.
Hint #1 In the case of 6-sided dice, if all the dice have opposite sides adding up to 7 (1 opposite to 6, 2 opposite to 5, 3 opposite to 4) then there are only 2 groups. Try to prove that result, then look at the case when opposite faces don't have to add up to 7. How many of those cases are there in total?
Hint #2 So, if you fix the top and bottom numbers, you get 2 groups (that's the result of hint #1 above). In how many ways can you fix the top and bottom numbers?
Best Answer
The thing with dice games is that we often want to know the odds of getting a particular event ('I have a pair of sixes').
Your result is correct for calculating the number of possible combinations, with interchangeable dices. However, if we want to find a result about, say, the sum of the dices, this way of counting cases is not particularly useful since there are less chances to get the combination (6,6,6,6,6) than the combination (1,2,3,4,5).
Your book calculates the number of 'outcomes' as if each dice is of a different colour (hence, not interchangeable) and getting (1,2,3,4,5) is not the same as getting (2,3,4,5,1). The main point of this approach is to have outcomes that all have the same odds of occuring (namely, $1/6^5$). Then you can solve a lot of probability questions (odds of at least one 6, odds of sum>20, etc) just by counting the favourable cases.