Expressing power sum symmetric polynomials in terms of lower degree power sums

Combining the trace formula proposed by Gjergji Zaimi and Qiaochu Yuan,
$$
p_k={rm Tr}begin{pmatrix} e_1 & 1 & cdots & 0 \
-e_2 & 0 & ddots & vdots \
vdots & vdots & ddots & 1 \
(-1)^{N-1}e_N & 0 & cdots & 0 end{pmatrix}^{k},
$$
with the formula quoted by Peter Erskin,
$$
e_n=frac1{n!} begin{vmatrix}p_1 & 1 & 0 & cdots\ p_2 & p_1 & 2 & 0 & cdots \ vdots&& ddots & ddots \ p_{n-1} & p_{n-2} & cdots & p_1 & n-1 \ p_n & p_{n-1} & cdots & p_2 & p_1 end{vmatrix},
$$
Mathematica produces the following expansions of $p_k$:

---

$$N=2$$

$$
p_3=-frac{1}{2} p_1^3+frac{3}{2} p_1p_2
$$

$$
p_4=-frac{1}{2} p_1^4+p_1^2p_2+frac{1}{2} p_2^2
$$

$$
p_5=-frac{1}{4} p_1^5+frac{5}{4} p_1p_2 ^2
$$

$$
p_6=-frac{3}{4} p_1^4p_2+frac{3}{2} p_1^2p_2^2+frac{1}{4} p_2^3
$$

$$
p_7=frac{1}{8} p_1^7-frac{7}{8} p_1^5p_2+frac{7}{8} p_1^3p_2^2+frac{7}{8} p_1p_2^3
$$

$$
p_8=frac{1}{8} p_1^8-frac{1}{2} p_1^6p_2-frac{1}{4} p_1^4p_2^2+frac{3}{2} p_1^2p_2^3+frac{1}{8} p_2^4
$$

$$
p_9=frac{1}{16} p_1^9-frac{9}{8} p_1^5p_2^2+frac{3}{2} p_1^3p_2^3
+frac{9}{16} p_1p_2^4
$$

$$
p_{10}=frac{5}{16} p_1^8p_2-frac{5}{4} p_1^6p_2^2+frac{5}{8} p_1^4p_2^3
+frac{5}{4} p_1^2p_2^4+frac{1}{16} p_2^5
$$

$$
p_{11}=-frac{1}{32} p_1^{11}+frac{11}{32} p_1^9p_2-frac{11}{16} p_1^7p_2^2-frac{11}{16} p_1^5p_2^3
+frac{55}{32} p_1^3p_2^4+frac{11}{32} p_1p_2^5
$$

---

$$N=3$$

$$
p_4=frac{1}{6} p_1^4-p_1^2p_2+frac{1}{2} p_2^2+ frac{4}{3} p_1p_3
$$

$$
p_5=frac{1}{6} p_1^5-frac{5}{6} p_1^3p_2+frac{5}{6} p_1^2p_3+frac{5}{6} p_2p_3
$$

$$
p_6=frac{1}{12} p_1^6-frac{1}{4} p_1^4p_2-frac{3}{4} p_1^2p_2^2+frac{1}{4} p_2^3+frac{1}{3} p_1^3p_2^3+p_1 p_2 p_3 +frac{1}{3} p_3^2
$$

$$
p_7=frac{1}{36} p_1^7-frac{7}{12} p_1^3p_2^2+frac{7}{36} p_1^4p_3+frac{7}{12} p_2^2p_3+frac{7}{9} p_1p_3^2
$$

$$
p_8=frac{1}{72} p_1^8-frac{1}{18} p_1^6p_2+frac{1}{12} p_1^4p_2^2-frac{1}{2} p_1^2p_2^3+frac{1}{8} p_2^4+frac{2}{9} p_1^5p_3
$$
$$
-frac{8}{9} p_1^3p_2p_3+frac{2}{3} p_1p_2^2p_3+frac{8}{9} p_1^2p_3^2+frac{4}{9} p_2p_3^2
$$

---

$$N=4$$

$$
p_5=-frac{1}{24} p_1^5+frac{5}{12} p_1^3p_2-frac{5}{8} p_1p_2^2-frac{5}{6} p_1^2p_3+frac{5}{6} p_2p_3+frac{5}{4} p_1p_4
$$

$$
p_6=-frac{1}{24} p_1^6+frac{3}{8} p_1^4p_2-frac{3}{8} p_1^2p_2^2-frac{1}{8} p_2^3-frac{2}{3} p_1^3p_3+frac{1}{3} p_3^2+frac{3}{4} p_1^2p_4+frac{3}{4} p_2p_4
$$

$$
p_7=-frac{1}{48} p_1^7+frac{7}{48} p_1^5p_2+frac{7}{48} p_1^3p_2^2-frac{7}{16} p_1p_2^2-frac{7}{24} p_1^4p_3-frac{7}{12} p_1^2p_2p_3
$$
$$+frac{7}{24} p_2^2p_3
+frac{7}{24} p_1^3p_4+frac{7}{8} p_1p_2p_4+frac{7}{12} p_3p_4
$$

$$
p_8=-frac{1}{144} p_1^8+frac{1}{36} p_1^6p_2+frac{5}{24} p_1^4p_2^2
-frac{1}{4} p_1^2p_2^2-frac{1}{16} p_2^4
$$
$$
-frac{1}{9} p_1^5p_3-frac{2}{9} p_1^3p_2p_3
-frac{1}{3} p_1p_2^2p_3-frac{4}{9} p_1^2p_3^2+frac{4}{9} p_2p_3^2
$$
$$
+frac{1}{12} p_1^4p_4+frac{1}{2} p_1^2p_2p_4+frac{1}{4} p_2^2p_4
+frac{2}{3} p_1p_3p_4+frac{1}{4} p_4^2.
$$

---

It seems to me that a nice and compact formula for $a_{k,rho}$ does exist. Indeed,
the coefficients in the above examples are extremely simple.

In particular, I observe that the last terms in each of $p_k$ for $N=8$
have the form
$$
k prod_j frac{1}{j^{r_j}r_j!}p_j^{r_j},
$$
which corresponds to
$$
a_{k,rho}=kprod_j frac{1}{j^{r_j}r_j!}.
$$
This formula (whose structure resembles the coefficients in the expansion of Schur functions quoted by Peter Erskin) also works for all terms of the type $p_jp_{k-j}$ at arbitrary $N$.

Apparently, this is not a general formula, as can be seen from the coefficients in front
of $p_1^k$ which do depend on $N$.
I believe, however, that the general formula for $a_{k,rho}$ with $N$ properly included should not be much more complex than the empirical one above.

Hope this helps.