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


[Dress and Siebeneicher](https://www.sciencedirect.com/science/article/pii/0001870889900273?via%3Dihub) 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](https://oeis.org/A263916) 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](https://www.ias.ac.in/article/fulltext/reso/015/12/1111-1121)" by Mihet and "[Revisiting Kummer's and Legendre's Formulae](https://www.ias.ac.in/article/fulltext/reso/010/02/0062-0071)" 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 $b\_n(a\_1,...,a\_n) = \frac{1}{n}[R\_n(b\_1-F.(a\_1,...),b\_2,b\_3,...,b\_{n-1},0] - R\_n(0,-b\_2,-2b\_3,...,-(n-2)b\_{n-1},0)].$



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


