I am a beginner to invariant theory and I am reading the Wikipedia page on the subject. There it says the following. Suppose $G = SL_n$ the group of $n*n$ matrices with complex coefficients, and $M_n$ is the vector space of $n*n$ matrices. Let the action of $G$ on $V$ be the normal by one by left multiplication. Then there is an induced action of $G$ on $\mathbb{C}[V]$ given by
$$g \cdot f (x) = f( g^{-1} x) $$ for all $g \in G, f \in \mathbb{C}[V]$, and $x \in V.$
We then consider the $\mathbb{C}-$algebra of invariant polynomials, that is,
$$\mathbb{C}[V]^G = \{ f \in \mathbb{C}[V] \mid g \cdot f = f, \, \forall g \in G \}.$$
The website claims that this is finitely generated by the determinant polynomial. I can understand intuitively why that should be so but don't know how to go about proving it. Any tips? Also, was this proven by Hilbert in a text somewhere else?
Here is a geometrically minded, intuitive proof. Note that any function on $M_n$ is determined by its restriction to $GL_n$, as this is dense. Then note that a polynomial function on $GL_n$ which is fixed by $SL_n$ is a function on the cosets $GL_n/SL_n$, but by the determinant map, this is just $\mathbb{G}_m=\mathbb{C}^*$. So we see the pullback of the coordinate function via this map to $\mathbb{C}^*$ (aka the determinant) generates the invariant functions on $M_n$ under the action of $SL_n$.
To make this totally rigorous, one needs to know a bit about quotients by algebraic groups/their rings of functions, but this is the intuition.