If $U_1,\ldots, U_n$ are i.i.d. uniform random variables, then I know that the order statistics satisfy $$(U_{(1)},\ldots, U_{(n)}) \overset{d}{=} \left(\frac{X_1}{\sum_{i=1}^{n+1} X_i}, \frac{X_1+X_2}{\sum_{i=1}^{n+1} X_i},\ldots, \frac{\sum_{i=1}^n X_i}{\sum_{i=1}^{n+1} X_i}\right),$$ where the $X_i$ are i.i.d. $\operatorname{Exponential}(1)$.
I know how to compute the expectation of some $U_{(k)}$ (it is $k/(n+1)$) by explicitly finding the pdf, but is there an elegant way to compute it using the reformulation involving exponential random variables? The following argument is suggestive, but incorrect. $$\mathbb{E}\frac{\sum_{i=1}^k X_i}{\sum_{i=1}^{n+1} X_i} \ne \frac{\mathbb{E}\sum_{i=1}^k X_i}{\mathbb{E}\sum_{i=1}^{n+1} X_i} = \frac{k}{n+1}.$$