Sum over permutations is 1

Yes. Write the sum as $$ \sum_{\pi \in S_n} n! \int_0^{\infty} e^{-\pi(1) x_1} \int_{0}^{\infty} e^{-(\pi(1)+\pi(2)) x_2} \ldots \int_0^{\infty} e^{-(\pi(1) + \ldots +\pi(n))x_n} dx_n \ldots dx_1. $$ Upon writing $y_n = x_n$, $y_{n-1}=x_n+x_{n-1}$, etc this becomes $$ \sum_{\pi \in {S_n} } n! \int_{y_1 \ge y_2 \ge \ldots y_n \ge 0} e^{-\sum_{i=1}^{n} \pi(i)y_i } dy_1 \ldots dy_n. $$ Since the sum is over all $\pi \in S_n$, it doesn't matter that we are in the region $y_1 \ge y_2 \ge \ldots \ge y_n$ -- for any other ordering of the non-negative variables $y_i$, the answer would be the same. So recast the above as $$ \int_0^{\infty} \ldots \int_0^{\infty} \sum_{\pi \in S_n} e^{-\sum_{i=1}^{n} \pi(i) y_i} dy_1 \ldots dy_n = \sum_{\pi \in S_n} \frac{1}{n!} = 1. $$

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.

Purely algebraic proof. We induct on $n$. Fix $\pi_n$. This part of the sum equals $z_{\pi_n}/(z_1+\dots+z_n)$ by induction proposition. Now sum up by all values of $\pi_n$.