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**.



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 
$$ 
\int\_0^{\infty} \ldots \int\_0^{\infty} \sum\_{\pi \in S\_n} e^{-\sum\_{i=1}^{n} \pi(i) y\_i} dy\_1 \ldots dy\_n 
= \sum\_{\pi \in S\_n} \frac{1}{n!} = 1.
$$


