TITLE: Is there a General Formula for the Transition Matrix from Products of Elementary Symmetric Polynomials to Monomial Symmetric Functions?
QUESTION [13 upvotes]: Given the elementary symmetric polynomials $e_k(X_1,X_2,...,X_N)$ generated via
$$
\prod_{k=1}^{N} (t+X_k) = e_0t^N + e_1t^{N-1} + \cdots + e_N.
$$
How can one get the monomial symmetric functions $m_\lambda(X_1,X_2,...,X_N)$ as products and sums in $e_k$?
For example: $N=4$ 
$$
m_{(2,1,1,0)}=X_1^2X_2X_3 + \text{all permutations}= e_3\cdot e_1 - 4 e_4 ,
$$
$$
m_{(2,2,0,0)}=X_1^2X_2^2 + ... = e_2^2-6e_4 - 2m_{(2,1,1,0)}=e_2^2 - 2e_3\cdot e_1 +2e_4
$$
It seems clear that the products on the RHS run over all partitions $\mu$ of $N$. For each $\lambda$ there should be a set of integers $C_{\lambda\mu}$ such that
$$
m_\lambda  = \sum_\mu C_{\lambda \mu} \prod_j e_{\mu_j} 
$$
is true. Putting all values of $c_{\lambda \mu}$ together, you get the following equation:
$$
\left(
\begin{matrix}
m_{4,0,0,0}\\
m_{3,1,0,0}\\
m_{2,2,0,0}\\
m_{2,1,1,0}\\
m_{1,1,1,1}\\
\end{matrix}
\right)=
\left(
\begin{matrix}
-4&+4&+2&-4&+1\\
+4&-1&-2&+1&0\\
+2&-2&+1&0&0\\
-4&+1&0&0&0\\
+1&0&0&0&0\\
\end{matrix}
\right)\left(
\begin{matrix}
e_{4}\\
e_3e_1\\
e_2^2\\
e_2e_1^2\\
e_1^4\\
\end{matrix}
\right)
.
$$
The transition matrix $C_4$ is symmetric. This also holds for $N=3$, where $C_3$ is 
$$
\left(
\begin{matrix}
+3&-3&+1\\
-3&+1&0\\
+1&0&0\\
\end{matrix}
\right)
$$
and for $N=2$ we get
$$
\left(
\begin{matrix}
-2&+1\\
+1&0\\
\end{matrix}
\right).
$$
The main question is, if there exists a general formula for the entries of the matrices?
For a given $N$, is it a kind of composition of matrices $C_{k<N}$ and some few other matrices? 
Beside that, the following is also of interest: 
The matrices so far are symmetric and their entries sum up to $0$. Is this true in general? Is this related to (conjugate) Young Tableaux?
Balls&Boxes
If we bring the matrix on the LHS we get:
$$
\left(
\begin{matrix}
0 &0&0&0&1\\
0&0&0&1&4\\
0&0&1&2&6\\
0&1&2&5&12\\
1&4&6&12&24\\
\end{matrix}
\right)
\left(
\begin{matrix}
m_{4,0,0,0}\\
m_{3,1,0,0}\\
m_{2,2,0,0}\\
m_{2,1,1,0}\\
m_{1,1,1,1}\\
\end{matrix}
\right)=
\left(
\begin{matrix}
e_{4}\\
e_3e_1\\
e_2^2\\
e_2e_1^2\\
e_1^4\\
\end{matrix}
\right)\tag{*}
.
$$
Maybe it's easier to interpret these values, since they are all positive, in terms of balls, that have to be put into boxes, according to the following rules :
You are given $N$ balls. Your balls are now divided into parts (and bless god that, this is a math forum :-) according to a partition $\mu$. These are the products of $e_k$'s.
Now you are asked to put the balls partition-by-partition into a $N$ boxes. It is not allowed to put more than 1 ball in a box for the current partition.
The goal is to achieve a certain distribution of balls among the boxes, given by $\lambda$. These are the $m_\lambda$.
Power Sums
Or would it help to express $e_k$ as power sums via Newton-Girard formulas?
Here is the worked out example:
$$
\left(
\begin{matrix}
m_{4,0,0,0}\\
m_{3,1,0,0}\\
m_{2,2,0,0}\\
m_{2,1,1,0}\\
m_{1,1,1,1}\\
\end{matrix}
\right)=
\left(
\begin{matrix}
0& 0& 0& 0& 1\\
 0& 0& 0& 1 &-1\\
0& 0& 1/2& 0& -1/2\\
0& 1/2& -1/2& -1& 1\\
1/24& -1/4 &+1/8 &1/3 &-1/4\\
\end{matrix}
\right)
\left(
\begin{matrix}
p_1^4\\
p_2p_1^2\\
p_2^2\\
p_3p_1\\
p_4^1\\
\end{matrix}
\right)
$$

REPLY [5 votes]: Beside the very nice answer by Marc, I found this Introduction to Symmetric Functions by Mike Zabrocki:

$$ e_\mu = \sum_\lambda B_{\lambda \mu} m_\lambda\tag{2.28} $$ 
  where $B_{\lambda \mu}$ is the number of matricies with entries in $\{0, 1\}$ whose column sum is $\mu$ and row sum is equal to $\lambda$.
  $$\dots$$
  $B_{\lambda \mu}$is the number of ways of filling the the Young diagram for the partition  with $\mu_1$ 1s, $\mu_2$ 2s, etc. that are strictly increasing in the rows and there is no restriction on the relationship between the values in the columns.

In a table summarising all possible transitions at the end of chapter 2.1 they also mention what I was previously looking for
$$ m_\mu = \sum_\lambda G_{\lambda \mu} e_\lambda, $$ 
but leave it as an exercise to determine some sort of formula for these ($G_{\lambda \mu}$) coefficients.