An identity involving partitions I have been trying to prove an identity using a combinatorial argument or another technique.  I need to define the following first:
$$R(n,m) = \{ (r_1,r_2,...,r_n): \sum_{i=1}^n r_i = m \text{ and } r_i \text{ is a positive integer }\}.$$
I believe there is a name for this set.  If so let me know.  I think of it as the set of ordered partitions of m of size n.
The identity I want to prove is the following.  For $m \ge 2$:
$$\sum_{n=1}^m \frac{(-1)^n}{n!}\sum_{r \in R(n,m)}\frac{1}{r_1r_2r_3...r_n}=0$$
The reason I know this identity is true is that the left side is the coefficient of $T^m$ in the power series expansion of $\exp (\log(T+1)) = T+1$ where exp and log are defined by $$\exp(T) = \sum_{n=0}^\infty \frac{T^n}{n!}$$ and $$\log(T+1)=-\sum_{n=1}^\infty(-1)^n\frac{T^n}{n}.$$  I am trying to use the identity to show that exp and log as defined by these power series are in fact inverses.  Thanks for any help.
 A: Note:  This question was answered also by some one else.
See the bottom for the link.
After much thought, I can answer my own question.   I find it easier to start from the beginning.  The initial problem I had was proving that if $$\exp(T) = \sum_{n=0}^{\infty}\frac{T^n}{n!}$$ and $$\log(1+T)=\sum_{n=1}^{\infty}\frac{(-1)^{k-1}T^k}{k}$$ then $$\exp(\log(1+T)) = 1+T$$
We have $$\exp(\log(1+T))=\frac{1}{0!}+\frac{(\frac{T}{1}-\frac{T^2}{2}+\ldots )}{1!}+\frac{(\frac{T}{1}-\frac{T^2}{2}+\ldots )^2}{2!}+\cdots $$
Collecting the first few terms of $\exp(\log(1+T))=\sum_{n=0}^{\infty}a_nT^n$ we get:
$$a_0 = \frac{1}{0!}=1$$
$$a_1T =\frac{T}{1!} = T$$
$$a_2T^2 = -\frac{T^2}{2\cdot 1!}+\frac{T^2}{1\cdot 1\cdot 2!}$$
$$a_3T^3 = \frac{T^3}{3\cdot 1!}-\frac{T^3}{1\cdot 2\cdot 2!}-\frac{T^3}{2\cdot 1\cdot 2!}+\frac{T^3}{1\cdot 1\cdot 1\cdot 3!}$$
$$a_4T^4=-\frac{T^4}{4\cdot 1!}+\frac{T^4}{1\cdot 3\cdot 2!}+\frac{T^4}{2\cdot 2\cdot 2!}+\frac{T^4}{3\cdot 1 \cdot 2!}-\frac{T^4}{1\cdot 1\cdot 2\cdot 3!}-\frac{T^4}{1\cdot 2\cdot 1\cdot 3!}-\frac{T^4}{2\cdot 1\cdot 1\cdot 3!}+\frac{T^4}{1\cdot 1\cdot 1\cdot 1\cdot 4!}$$
To find a general formula for $a_n$ we let $R(m,n)$ be the set of m-tuples $(r_1,r_2,\ldots,r_m)$ of positive integers for which $\sum r_i=n$.  Then for $n\ge 1$,
$$a_n=-\sum_{m=1}^{n}(-1)^{n-m}\sum_{r\in R(m,n)}\frac{1}{(\Pi {r_i})m!}$$
I want to show that for $n\ge 2$, we have $a_n=0$.  (This is equivalent to what I was asking in my original question above.)  Now we will associate each m-tuple in $R(m,n)$ with a multi-set.  So $(r_1,r_2,\ldots,r_m)$ is associated with the multi-set $\{r_1,\ldots, r_m\}$.  Let the multi-set $K_M=\{k_1,k_2,\ldots, k_l\}$ be the multiplicities of the multi-set $M$.
Let $\mathcal M(m,n)$ be the collection of all multi-sets of $m$ positive integers whose sum is $n$.  Let $M\in \mathcal M(m,n)$ with $K_M = \{k_1,k_2,\ldots,k_l\}$. There are $$\frac{m!}{k_1!k_2!\ldots k_l!}$$ m-tuples in $R(m,n)$ associated with this multi-set.
Note: You may have seen this formula in a discrete math class.  Does anyone remember how to find the number of ways to arrange the letters of in the word MISSISSIPPI?  It's the same idea here.  The letters form a multi-set.  An arrangement is an m-tuple.
Knowing this, it is now possible to write $a_n$ as the sum over multi-sets instead as:
$$a_n=-\sum_{m=1}^{n}(-1)^{n-m}\sum_{M\in \mathcal M(m,n)}\frac{m!}{(\Pi_{k\in K_M}(k!)}\frac{1}{(\Pi r_i)m!}$$
Which can be rewritten as
$$a_n=\frac{-1}{n!}\sum_{m=1}^{n}(-1)^{n-m}\sum_{M\in \mathcal M(m,n)}\frac{n!}{(\Pi_{k\in K_M}k!)(\Pi r_i)}$$
Separating the negative and positive terms of this sum we get:
$$a_n=\frac{-1}{n!}(\sum_{M\in \mathcal M_e}\frac{n!}{(\Pi_{k\in K_M}k!)(\Pi r_i)}-\sum_{M\in \mathcal M_o}\frac{n!}{(\Pi_{k\in K_M}k!)(\Pi r_i)})$$
where $\mathcal M_e$ and $\mathcal M_o$ are the multisets (adding to n) for which n-m is even and n-m is odd respectively.
We can identify each multi-set $M$ of positive integers, which add to n, with a conjugacy class of $S_n$ by setting $M$ as the cycle type of the conjugacy class.  For example, $M=\{2,3,2,1,1,4\}$ can be identified with the class $H_M\subset S_{13}$ of all permutations of the form
$$\phi = (ab)(cde)(fg)(h)(i)(jklm)$$
where the product is made up of disjoint cycles.  Note that any cycle of length r can be written in r different ways and we can interchange the position in the product of two cycles of the same length.  For example,
$$\phi = (ab)(cde)(fg)(h)(i)(jklm)=(fg)(cde)(ab)(i)(h)(lmjk)$$
so the multiset $M=\{2,3,2,1,1,4\}$ is identified with the conjugacy class $H_M$ and $H_M$ has cardinality
$$\frac{13!}{2!1!2!1!\cdot 2\cdot 3\cdot 2\cdot 1\cdot 1\cdot 4}=\frac{n!}{(\Pi_{k\in K_M}k!)(\Pi r_i)}$$
For any given multi-set $M\in \mathcal M(m,n)$, every $\phi\in H_M$ can be written as the product of transpositions (cycle of length 2).  A cycle of length $r_i$ is the product of $r_i-1$ transpositions.  Thus $\phi$ is the product of $\sum_{i=1}^{m}(r_i-1)= n-m$ transpositions.  If $n-m$ is even then $\phi\in A_n$ so $H_M\subset A_n$.  If $n-m$ is odd then $H_M\subset S_n\setminus A_n$.  Both $A_n$ and $S_n\setminus A_n$ are half the cardinality of $S_n$, for $n\ge 2$.  We therefore have:
$$a_n=\frac{-1}{n!}(\sum_{M\in \mathcal M_e}|H_M|-\sum_{M\in \mathcal M_o}|H_M|$$
Thus,
$$a_n=\frac{-1}{n!}(|A_n|-|S_n\setminus A_n|)=0$$
for $n\ge 2$.
I didn't notice this before but this question was answered quite nicely here:
Composition of formal power series of exp and log
