= Solution
Let $R=\mathbb C[V]=\operatorname{Sym}(V^*)$, with its degree grading, and let $A=R^G$ be the <polynomial invariant ring>. For a <compact group> average polynomials over $G$; for a complex reductive group average over its Zariski-dense compact form from part (i). Each degree piece is finite dimensional, so this defines a polynomial and gives the <Reynolds operator on a polynomial invariant ring>
$$
\mathcal R:R\to A,\qquad\mathcal R|_A=\operatorname{id},\qquad\mathcal R(af)=a\mathcal R(f)\quad(a\in A).
$$
It preserves degree. For the reductive case compact invariance is full group invariance by Zariski density.
Let $A_+$ be the positive-degree invariants and form the ideal $I=RA_+$. The <Hilbert basis theorem> makes $R$ a <Noetherian ring>, so finitely many homogeneous invariants $f_1,\ldots,f_s\in A_+$ generate $I$ as an $R$-ideal. Indeed start with a finite ideal-generating set and collect the finitely many invariants occurring in expressions for those generators; then replace them by homogeneous pieces.
We claim that these same $f_i$ generate $A$ as an algebra. For a homogeneous invariant $f$ of degree $d>0$, write $f=\sum_i a_i f_i$ with homogeneous coefficients of degree $d-\deg f_i$; terms of negative degree are absent. Apply the <Reynolds operator on a polynomial invariant ring> to get $f=\sum_i\mathcal R(a_i)f_i$. The averaged coefficients are invariant and have smaller degree than $d$. Starting from the constants, induction on degree puts them and then $f$ in $\mathbb C[f_1,\ldots,f_s]$. Taking homogeneous pieces covers every invariant. This proves <Hilbert finite generation for reductive invariants>:
$$
\boxed{\mathbb C[V]^G=\mathbb C[f_1,\ldots,f_s].}
$$
Noetherianity alone would not suffice for a subring; the Reynolds module property is the step that makes the induction work.
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