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?

On the 'ubiquity' of the Faber polynomials:

These Faber partition polynomials and their associated Faber Appell polynomials with the shady relation $FA_n(x) = (x + F.)^n$ crop up in mutitude 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 $b_n(a_1,...,a_n) = \frac{1}{n}[R_n(b_1-F.(a_1,...),b_2,b_3,...,b_{n-1},0] - R_n(0,-b_2,-2b_3,...,-(n-2)b_{n-1},0)].$