I just had to make use of an elementary rational function identity (below). The proof is a straightforward exercise, but that isn't the point. First, "my" identity is almost surely
not original, but I don't have a reference for it. Perhaps someone knows it (like a lost cat without a collar) or, more likely, could spot this as a special case of a more general identity. Second, the obvious proof is not much of an explanation: a combinatorial identity often arises for a conceptual reason, and I'd be happy to hear if anyone sees mathematics behind this one.
Let $f(x_1,\ldots,x_n)=\prod_{p=1}^n\big(\sum_{i=p}^n x_i\big)^{-1}$. Then
$$
f(x_1,\ldots,x_n)+f(x_2,x_1,x_3,\ldots,x_n)+\cdots+f(x_2,\ldots,x_n,x_1)=\big(\sum_{i=1}^n x_i\big)/x_1\cdot f(x_1,\ldots,x_n),
$$
where $x_1$ appears as the $i$th argument to $f$ in the $i$th summand on the left side, for $1\leq i\leq n$. But why?
Best Answer
I have seen a cat of a similar breed in the representation theory of symmetric groups. Out of habit, let me quote a lemma attributed to Littlewood in
Donald Knutson, $\lambda$-rings and the Representation Theory of the Symmetric Group, Springer 1973 (LNM #308), Chapter III, section 2, p. 149:
$\sum\limits_{\sigma\in S_n} f\left(x_{\sigma\left(1\right)},x_{\sigma\left(2\right)},...,x_{\sigma\left(n\right)}\right) = \frac{1}{x_1x_2...x_n}$.
At the moment, neither does this cat imply yours, nor the other way round. But can we cross them?
Let me try. The left paw side of your cat is $\sum\limits_{\sigma\in \mathrm{Sh}\left(1,n-1\right)} f\left(x_{\sigma^{-1}\left(1\right)},x_{\sigma^{-1}\left(2\right)},...,x_{\sigma^{-1}\left(n\right)}\right)$, where $\mathrm{Sh}\left(a,b\right)$ is defined as the subgroup
$\left\lbrace \sigma \in S_{a+b} \mid \sigma\left(1\right) < \sigma\left(2\right) < ... < \sigma\left(a\right) \text{ and } \sigma\left(a+1\right) < \sigma\left(a+2\right) < ... < \sigma\left(a+b\right) \right\rbrace$
of the symmetric group $S_{a+b}$. (The elements of this subgroup $\mathrm{Sh}\left(a,b\right)$ are known as $\left(a,b\right)$-shuffles.) Now I suspect tat
$\sum\limits_{\sigma\in \mathrm{Sh}\left(a,b\right)} f\left(x_{\sigma^{-1}\left(1\right)},x_{\sigma^{-1}\left(2\right)},...,x_{\sigma^{-1}\left(a+b\right)}\right) = f\left(x_1,x_2,...,x_a\right) f\left(x_{a+1},x_{a+2},...,x_{a+b}\right)$
for any $a$ and $b$ and any $x_i$.
This generalizes your cat. Does it generalize Littlewood's? Yes, at least if we generalize it even further, to the so-called $\left(a_1,a_2,...,a_k\right)$-multishuffles (which are permutations $\sigma\in S_{a_1+a_2+...+a_k}$ increasing on each of the intervals $\left[a_i+1,a_{i+1}\right]$, where $a_0=0$ and $a_{k+1}=n$). This is not much of a generalization, since it follows from the $\left(a,b\right)$-shuffle version by induction over $k$, but applying it to $\left(1,1,...,1\right)$-multishuffles (which are simply all the elements of $S_n$) yields Littlewood's cat.
Now I see that Littlewood's cat even follows from yours, if we notice that every permutation $\sigma\in S_n$ can be written uniquely as a product $t_1t_2...t_{n-1}$, where each of the $t_k$ moves the $k$ some places to the right. (This is one of the stupid sorting algorithms.)
Oh, and I don't have a proof of my cat, but it can catch mice, so it's a good cat, isn't it?