The extraction (orthogonality) identity

First note the orthogonality relation $\int_0^1 e(m\alpha)\, d\alpha$$=1$ if $m=0$, and $=0$ if $m\neq 0$: writing $e(m\alpha)=\cos(2\pi m\alpha)+i\sin(2\pi m\alpha)$, for $m\neq0$ this is a full number of periods of a sine/cosine wave over $[0,1)$, which integrates to $0$; for $m=0$ the integrand is the constant $1$.

Now expand the $k$-th power of the generating function: $F_A(\alpha)^k=\Big(\sum_{a\in A}e(a\alpha)\Big)^k=\sum_{(a_1,\dots,a_k)\in A^k} e\big((a_1+\cdots+a_k)\alpha\big)$, a finite sum over all ordered $k$-tuples from $A$, grouped by the exponent $m=a_1+\cdots+a_k$.

Multiply both sides by $e(-n\alpha)$ and integrate term by term over $[0,1)$ (legitimate because the sum is finite, so integration and summation commute): $\int_0^1 F_A(\alpha)^k e(-n\alpha)\,d\alpha=\sum_{(a_1,\dots,a_k)\in A^k}\int_0^1 e\big((a_1+\cdots+a_k-n)\alpha\big)\,d\alpha$.

By the orthogonality relation, each summand on the right is $1$ exactly when $a_1+\cdots+a_k=n$ and $0$ otherwise. So the whole sum collapses to exactly the count of tuples with $a_1+\cdots+a_k=n$, i.e. $r_k(n)$. This proves the identity exactly, with no approximation involved: it is an identity, not an asymptotic.