TITLE: Sum over permutations is 1
QUESTION [28 upvotes]: This might be easy, but let's see.

Question 1. If $\mathfrak{S}_n$ is the group of permutations on $[n]$, then is the following true?
  $$\sum_{\pi\in\mathfrak{S}_n}\prod_{j=1}^n\frac{j}{\pi(1)+\pi(2)+\cdots+\pi(j)}=1.$$

Update. After Lucia's answer, I got motivated to ask:

Question 2. Let $z_1,\dots,z_n$ be indeterminates. Is this identity true too?
  $$\sum_{\pi\in\mathfrak{S}_n}\prod_{j=1}^n\frac{z_j}{z_{\pi(1)}+z_{\pi(2)}+\cdots+z_{\pi(j)}}=1.$$

Update. Now that we've an analytic and an algebraic proof, is there a combinatorial argument too? At least for Question 1.

REPLY [28 votes]: Here is a probabilistic, or, if you wish, combinatorial proof. Assume that we have $n$ baskets containing $z_1,\dots,z_n$ balls respectively (well, let them be positive integers). Choose a random ball (all balls have equal probability) and forbid all the balls from its basket. Repeat with $n-1$ remaining baskets and so on. What is the probability that we consecutively get balls from baskets $\pi_n,\pi_{n-1},\dots,\pi_1$? Yes, it equals $$\prod_{j=1}^n\frac{z_{\pi_j}}{z_{\pi(1)}+z_{\pi(2)}+\cdots+z_{\pi(j)}}$$
(we start multiplying from $j=n$ downto $j=1$). The sum of all these probabilities equals 1.