A Newton identity and the primes--the Faber partition polynomials and modular arithmetic

Question

[Edit, July 6, 2022: Removed erroneous characterization of Faber polynomials as an Appell sequence.]

Dress and Siebeneicher in their tale of the Burnside family express an opinion (1.2) that, if I read it correctly, leads me to believe that the classic Faber partition polynomials which satisfy

$\ln[A(x)] = \ln[1 + a_1 x + a_2 x^2 + \cdots ] = \sum_{\geq 1} -F_n(a_1,...,a_n)\; \frac{x^n}{n}$ have the property

$[F_n(a_1,a_2,...,a_n)- (-a_1)^n] \; mod(n) \; = 0$

for $n$ prime and integral indeterminates $a_n$ .

I've been assured by a reputable authority that such is obvious. Can someone provide an 'obvious' proof or a least some other published hearsay on this?

Gessel and Ree in "Lattice paths and Faber polynomials" give (p. 4) a multinomial- coefficient type of expression for the coefficients of the Faber polynomials $FP_n(u)$ for which $FP_n(0)=F_n[a_1,a_2,...,a_n]$. (G & R use the notation $F_n(u)$ for what I denote as $FP_n(u)$.)

[Edit, July 6, 2022: Motivated by Peter's answer, I found two nice intros for the uninitiated to Kummer's theorem--"Legendre’s and Kummer’s Theorems Again" by Mihet and "Revisiting Kummer's and Legendre's Formulae" by Sury.]

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On the 'ubiquity' of the Faber polynomials and Faber partition polynomials:

These Faber polynomials and their associated Faber partition polynomials crop up in multitude of discussions: in symmetric function theory in the Newton-Girard-Waring identities expressing the power symmetric polynomials in terms of the elementary symmetric polynomials; in operational calculus for a generic raising operator for Appell polynomials; in complex function theory in extending an analytic function defined on a closed curve to an analytic function within the curve (providing harmonic functions with prescribed boundary conditions); in an analog of Fourier series expansions of complex functions; in extracting the numerical values of the indeterminates of compositional partition polynomials given numerical evaluations of those polynomials; in relations between determinants and traces; in algebraic/geometric K-theory; in the combinatorial/analytic properties of random walks, lattice paths, noncrossing partitions, and associahedra; and in determining the compositional inverse of certain Laurent series--in fact, in an orgy with the family of reciprocal polynomials $R_n$ birthed by $1/A(x) = \sum_{n \geq 0} R_n(a_1,...,a_n) x^n$, the compositional inverse $L^{(-1)}(z) = z + b_1 +b_2/z +b_3/z^2 + \cdots $ of the formal Laurent series $L(z) =z + a_1 +a_2/z +a_3/z^2 + \cdots$ is given by the umbral recursion formula $b_n(a_1,...,a_n) = \frac{1}{n}[R_n(b_1-F.(a_1,...),b_2,b_3,...,b_{n-1},0] - R_n(0,-b_2,-2b_3,...,-(n-2)b_{n-1},0)].$

Answer 1

From $$\ln[A(x)] = \ln[1 + a_1 x + a_2 x^2 + \cdots ] = \sum_{n \geq 1} -F_n(a_1,...,a_n)\; \frac{x^n}{n}$$ and letting $A'(x) = A(x) - 1$ we have $$\begin{eqnarray*} F_n(a_1,...,a_n) &=& n [x^n] \sum_{i=1}^{\infty} \frac{(-A'(x))^i}{i} \\ &=& (-a_1)^n + n [x^n] \sum_{i=1}^{n-1} \frac{(-A'(x))^i}{i} \\ %&=& n \sum_{\lambda \, \vdash \, n} \frac{1}{\operatorname{len}(\lambda)} \binom{\operatorname{len}(\lambda)}{f_1, \ldots, f_n} \prod_i (-a_i)^{f_i} \\ &=& (-a_1)^n + n \sum_{\lambda \, \vdash \, n,\; \lambda \neq 1^n} \frac{1}{\operatorname{len}(\lambda)} \binom{\operatorname{len}(\lambda)}{f_1, \ldots, f_n} \prod_j (-a_j)^{f_j} \\ \end{eqnarray*}$$

where the sum in the last line is over partitions $\lambda = 1^{f_1} 2^{f_2} \cdots n^{f_n}$ with $\sum_i if_i = n$ and the length $\operatorname{len}(\lambda)$ defined as $\sum_i f_i$, and the partition $\lambda = 1^n$ giving $(-a_1)^n$ is pulled out of the sum.

It remains to show that all of the multinomial coefficients in $[x^n](-A'(x))^i$ with $i < n$ must be divisible by $i$; they're certainly not divisible by any prime greater than $i$, which by hypothesis includes $n$.

Let $p$ be a prime factor of $\operatorname{len}(\lambda) < n$. There must be some $c$ for which $f_c \neq 0 \pmod p$, since otherwise $p \mid n$ contradicting the primality of $n$. But then $$\binom{\operatorname{len}(\lambda)}{f_1, \ldots, f_n} = \binom{\operatorname{len}(\lambda)}{f_c} \binom{\operatorname{len}(\lambda) - f_c}{\{f_j : j \neq c\}}$$ and by Kummer's theorem $\nu_p \left( \binom{\operatorname{len}(\lambda)}{f_c} \right) \ge \nu_p \left(\operatorname{len}(\lambda)\right)$. Therefore the multinomial coefficient is indeed divisible by $\operatorname{len}(\lambda)$, giving the desired result.