$X_1,X_2,...,X_n$ independent random variables are uniformly distributed on $[0,1]$.
So, we say that $P(X_1<X_2<...<X_n)=\frac{1}{n!}$
But why is that the correct answer? How do we calculate it?
According to the answer, my guess is: This case is like arranging $n$ people in a row. we have $n!$ permutations for that, and only one satisfies the requirement. Thus, we get $\frac{1}{n!}$.
Is that the way of thinking that should be?
There is zero probability that two variables are equal. We can forget this case of zero measure.
Take a permutation of indices $\sigma$. Consider $p(\sigma)$ the probability that the variables are ordered accordingly.
By symmetry $p(\sigma)$ does not depend on $\sigma$ and the sum over all permutations is $1$, hence the conclusion that $p(\sigma)=1/n!$.
It does not even matter that the variables are uniformly distributed (e.g. could be normally distributed R.V.). The symmetry argument always applies.
Essentially the argument is very similar to what to propose, only one has to reframe it into more formal probabilistic terms.