Ordered statistics, $\mathrm{E}(F(X_{(n)})-F(X_{(1)}))$, $F(x)$ continous common distribution

Question

From An Intermediate Course in Probability by Allan Gut:

Let $X_{1},X_{2},\ldots, X_{n}$ be independent, continuous random variables with common distribution function $F(x)$, and consider the order statistic $(X_{(1)},X_{(2)},\ldots, X_{(n)})$. Compute $\mathrm{E}(F(X_{(n)})-F(X_{(1)}))$.

The answer should be $\frac{n-1}{n+1}$. How to arrive at this answer?

$\mathrm{E}(X_{n}-X_{1})=\frac{n-1}{n+1}$ for $X\in U(0,1)$, but nothing is known about the distribution of $F(x)$, exept it is continous i.i.d.

Answer 1

Would you believe $F_X(X)\sim\mathcal U[0,1]$ ?

$$\begin{align}\mathsf P(F_X(X)\leq x)&=\mathsf P(X\leq F_X^{-1}(x))\\[1ex]&=F_X\circ F^{-1}_X(x)\\[1ex]&=x\mathbf 1_{x\in[0,1]}+\mathbf 1_{x\in(1,\infty)}\\[3ex]\therefore~F_X(X)&\sim\mathcal U[0,1]\end{align}$$