TITLE: A sum over partitions involving "subpartitions"
QUESTION [16 upvotes]: Consider the following sum over partitions of $n$:
$$ S(n)=\sum_{\substack {j_1,\dots,j_n\geq 0\\j_1+2j_2+\dots+nj_n=n}} \prod_{t=1}^n \frac{1}{j_t!t^{j_t}}f_t(j_1,\dots,j_t),$$
where
$$ f_t(j_1,\dots,j_t)=\begin{cases}\frac{j_t}{j_t+1}\,&\textrm{if } j_1+2j_2+\dots+(t-1)j_{t-1}=t-1 \\1 \,&\textrm{otherwise}\end{cases}.$$
I have strong numerical evidence that
$$ S(n)=\frac{1}{n+1},$$
but I cannot prove it, I was wondering if anyone could give me ideas.

Some observations: I wasn't able to compute it using a simple generating function, since I would need $f_t(j_1,\dots,j_t)$ to be a function only of $j_t$.
The function $$f(j_1,\dots, j_n)=\prod_{t=1}^n f_t(j_1,\dots,j_t)$$ can also be intepreted as a function of the cycle structure of a permutation depending on which "subpartitions" $j_1,\dots, j_n$ contains, or on the invariant subsets of $\{1,\dots, n\}$ under the permutations with the cycle structure given by the partition. $S(n)$ can be seen as the average of $f$ over the permutation group, since the $\prod_{t=1}^n \frac{1}{j_t!t^{j_t}}$ is the probability of drawing a permutation with $j_k$ $k-$cycles in its decomposition.
Ideally, I would like to compute, or at least bound
$$ S(n,x)=\sum_{\substack {j_1,\dots,j_n\geq 0\\j_1+2j_2+\dots+nj_n=n}} \prod_{t=1}^n \frac{x^{j_t}}{j_t!t^{j_t}}f_t(j_1,\dots,j_t),$$
for $x>0$, but I don't have a good conjecture on the form of this sum, except for $x=1$.

REPLY [2 votes]: This is just an extended comment providing an alternative view at $S(n)$, which (I hope) may lead to a solution.
Notice that $f_t(j_1,j_2,\dots,j_t) = 1-\frac{\delta_{j,t}}{1+j_t}$,
where
$$\delta_{j,t} := \big[j_1+2j_2+\dots+(t-1)j_{t-1} = t-1\big]$$
is an Iverson bracket.
Let $\bar n:=\{1,2,\dots,n\}$ and for a given partition exponents $j=(j_1,\dots,j_n)$ with $j_1 + 2j_2+\dots+nj_n=n$, define $J(j) := \{t\in\bar n\,:\,\delta_{j,t}=1\}$. Then
\begin{split}
(n+1)! S(n)&=(n+1)!\sum_{j:\ j_1 + 2j_2+\dots+nj_n=n}  \prod_{t=1}^n \frac{1}{j_t!t^{j_t}} \sum_{T\subseteq J(j)} (-1)^{|T|} \prod_{t\in T} \frac1{j_t+1} \\
&= \sum_{T\subseteq \bar n} (-1)^{|T|} \sum_{j:\ j_1 + 2j_2+\dots+nj_n=n\atop J(j)\supseteq T} \prod_{t=1}^n \frac{(n+1)!}{(j_t + [t\in T])!\,t^{j_t}}.
\end{split}
The last formula may be viewed as application of the inclusion-exclusion principle under a suitable combinatorial interpretation of its terms, which then would likely imply the needed $(n+1)!S(n)=n!$ out of the box. Unfortunately, I was not able to find such an interpretation so far, but have a gut feeling it's out there.