Expressing power sum symmetric polynomials in terms of lower degree power sums Is there an explicit formula expressing the power sum symmetric polynomials 
$$p_k(x_1,\ldots,x_N)=\sum\nolimits_{i=1}^N x_i^k = x_1^k+\cdots+x_N^k$$ 
of degree $k$ in $N < k$ variables entirely through the power 
sum symmetric polynomials $p_j(x_1,\ldots,x_N)$ of degrees $ j \le N $? 
Examples: 
$$N=1,\ k=2: \quad p_2=x^2=x\times x=p_1^2$$
$$N=2,\ k=3: \quad
p_3 = x^3 + y^3 = [3(x^2+y^2)(x+y)-(x+y)^3]/2 = (3 p_2 p_1-p_1^3)/2$$

What is the general formula?

I am looking for a formula similar to that for the expansion of the Schur functions
 $s_\lambda$ in terms of the symmetric power sums:
$$ s_\lambda=\sum_{\rho=(1^{r_1},2^{r_2},3^{r_3},\dots)}\chi^\lambda_\rho \prod_j \frac{p^{r_j}_j}{r_j!},$$
where the coefficients $\chi^\lambda_\rho$ are the characters of the representation of the symmetric group indexed by the partition $\lambda$ evaluated at elements of cycle type indexed by the partition $\rho=(1^{r_1},2^{r_2},3^{r_3},\dots)$, which contains 
$ r_j $ 
parts of length $j$. 
Clearly, the power sums of degree higher than $N$ can be expanded in a similar manner:
$$
p_k=\sum_{\rho}a_{k;\rho}\prod_{j=1}^N p_j^{r_j},
$$
where $\rho=(1^{r_1},2^{r_2},\dots,N^{r_N})$ is the partition of $k$ such that
$k=r_1+2r_2+3r_3+...+Nr_N$. 
In the above example for $N=2,\ k=3$ one has $a_{3;\ (1^{1},2^{1}) }=3/2$ and 
$a_{3;\  (1^{3},2^{0})}=-1/2$. 

My question can be thus reformulated as follows:
  given $r_1,...,r_N$ what is the
  explicit formula for $a_{k;\rho}$?


Note Added
Actually, Wikipedia tells us how to construct a certain explicit formula for $p_k$.
It gives the following expressions for $p_n$ with $n=N$ in terms of $ e_j, $
$$
p_n =
\begin{vmatrix}
e_1 & 1 & 0 & \cdots & \\\ 
2e_2 & e_1 & 1 & 0 & \cdots & \\\
3e_3 & e_2 & e_1 & 1 & \cdots & \\\
\vdots &&& \ddots & \ddots  & \\\
ne_n & e_{n-1} & \cdots & & e_1 &
\end{vmatrix},
$$
and for $e_n$ with $n=N$ in terms of $ p_j, $
$$
e_n=\frac1{n!}
\begin{vmatrix}p_1 & 1 & 0 & \cdots\\\ p_2 & p_1 & 2 & 0 & \cdots  \\\ \vdots&& \ddots & \ddots \\\ p_{n-1} & p_{n-2} & \cdots & p_1 & n-1 \\\ p_n & p_{n-1} & \cdots & p_2 & p_1
\end{vmatrix}.
$$
As far as I can see from the derivation described in Wikipedia, these determinant expressions are also valid for $p_n$ with $ n > N $ and for $e_n$ with $ n < N $.
For $p_n$ with $n>N$ one should take into account that all $ e_k=0 $ for $ k > N $, so that the resulting matrix has zeros in both right-upper and left-lower corners. 
Substituting the determinants for $e_j$ into the determinant for $p_k$, one gets the 
explicit formula which seems to solve the problem. 
However, I still don't know how to obtain the coefficients $a_{k;\rho}$ in the expansion of $ p_k $ in terms of the first $N$ power sums which would be the desired (really explicit) formula. 
 A: Assuming you have $n$ variables then for $k\geq n$, Robin Chapman's identity above can be written as
$$(p_n,p_{n-1},\dots, p_1)\begin{pmatrix}
  e_1 & 1 & \cdots & 0 \\\
  -e_2 & 0 & \ddots & \vdots \\\
  \vdots  & \vdots  & \ddots & 1  \\\
  (-1)^{n-1}e_n & 0 & \cdots & 0
 \end{pmatrix}^{k-n}=(p_k,p_{k-1},\dots, p_{k-n+1})$$
Now to finish the job you need to express the $e_i$'s in terms of the power sum symmetric functions too. This is given by $$e_n=\sum_{|\lambda|=n}(-1)^{|\lambda|-l(\lambda)} z_{\lambda}^{-1}p_{\lambda}$$ where $|\lambda|$ is the size of the partition $\lambda$ and $l(\lambda)$ is its length, $p_{\lambda}=p_{\lambda_1}p_{\lambda_2}\cdots$ and 
$$z_{\lambda}=\prod_{i\geq 1}\left(i^{m_i}\cdot m_i!\right)$$
where $m_i$ is the number of parts of $\lambda$ equal to $i$.

I thought I'd remark that the formulas you are quoting are all valid in $\Lambda_{\mathbb{Q}}$, the ring of symmetric functions in infinitely many variables while the one you are searching for is not, because the $p_\lambda$'s are an orthogonal basis in this ring with $\langle p_{\lambda},p_{\mu}\rangle =\delta_{\lambda \mu}z_{\lambda}$. This is also the same reason why the formula for Schur polynomials may contain arbitrary $p_{\lambda}$'s in it. In fact the reason why that formula is important is because it gives you the transition from the basis of Schur polynomials to that of power sum polynomials in $\Lambda_{\mathbb{Q}}$.
A: If $e_1,\ldots,e_N$ are the elementary symmetric functions of $x_1,\ldots,x_N$
then for $k\ge N$ one has
$$p_k=\sum_{j=1}^N(-1)^{j-1}e_j p_{k-j}.$$
This formula uses the elementary symmetric functions, which I presume you want to avoid,
but it means that for $k\ge 2N$ the $(N+1)$ by $(N+1)$ matrix
$$M_k=(p_{k-i-j})_{i,j=0}^N$$
has the null-vector $(1,-e_1,e_2,-e_3,\ldots,\pm e_N)$ and so $\det(M_k)=0$.
Expanding this out gives an explicit formula for $p_k$ as a rational function
(alas!) of $p_{k-1},\ldots,p_{k-2N}$.
Added I suppose one can express the $e_j$ in terms of $p_1,\ldots,p_n$
and put them into the above linear recurrence for $p_k$.
A: $$ p_k = -k\int_0^{2\pi}\frac{d\theta}{2\pi}e^{-i k \theta}
\ln\left\{ \int_0^{2\pi}\frac{d\phi}{2\pi}\ \sum_{m=0}^N e^{i m (\theta-\phi)}
\exp\left[ -\sum_{s=1}^N \frac{p_s}{s}\ e^{i s \phi}
\right]   \right\}. $$
