Probability of first increase in ordering of iid random variables

What is the probability that the first $n-1$ terms of iid Unif(0,1) random draws are in decreasing order, but the first n terms are not?

I know that due to exchangability, $\mathbb{P}({X}_{1}<{X}_{2}<\cdots <{X}_{n-1})=\frac{1}{n-1!}$. Why can't a similar exchangability argument hold to show that $\mathbb{P}({X}_{1}>{X}_{2}>\cdots >{X}_{n-1}<{X}_{n})=\frac{1}{n!}$? My reasoning is that the distribution of $({X}_{1},{X}_{2},\dots ,{X}_{n})$ is symmetric, so no matter which permutation of the random variables exist in the random vector, they will all have the same probability.

What is the probability that the first $n-1$ terms of iid Unif(0,1) random draws are in decreasing order, but the first n terms are not?

I know that due to exchangability, $\mathbb{P}({X}_{1}<{X}_{2}<\cdots <{X}_{n-1})=\frac{1}{n-1!}$. Why can't a similar exchangability argument hold to show that $\mathbb{P}({X}_{1}>{X}_{2}>\cdots >{X}_{n-1}<{X}_{n})=\frac{1}{n!}$? My reasoning is that the distribution of $({X}_{1},{X}_{2},\dots ,{X}_{n})$ is symmetric, so no matter which permutation of the random variables exist in the random vector, they will all have the same probability.