I think it's a matter of basic linear algebra.

Generally, if $A\in\mathrm{GL}_n\left(k\right)$ for some infinite field $k$, then $A^{-1}$ lies in the Zariski closure of the set $\left\lbrace 1,A,A^2,A^3,...\right\rbrace$.

*Proof.* Let $P\in k\left[\mathrm{M}_n\left(k\right)\right]$ be a polynomial such that $P\left(A^i\right)$ for every $i\in\mathbb N$. We must then prove that $P\left(A^{-1}\right)=0$ as well. Here we naively consider $k\left[\mathrm{M}_n\left(k\right)\right]$ as the algebra of all polynomial functions from $\mathrm{M}_n\left(k\right)$ to $k$ (since $k$ is infinite, this is, of course, isomorphic to the non-naive algebra of coordinate functions).

Define a $k$-algebra $U$ by $U=\bigoplus\limits_{i=0}^N \left(\mathrm{M}_n\left(k\right)\right)^{\otimes i}$ as a vector space, but with the multiplication being inherited from $\mathrm{M}_n\left(k\right)$ on each summand. So, as a vector space $U$ is a "cropped" tensor algebra over $\mathrm{M}_n\left(k\right)$, but as an algebra it is a direct product!

Let $N=\deg P$. Then the polynomial $P:\mathrm{M}_n\left(k\right)\to k$ can be written as $P=p\circ s$, where $U=\bigoplus\limits_{i=0}^N \left(\mathrm{M}_n\left(k\right)\right)^{\otimes i}$, where $s:\mathrm{M}_n\left(k\right)\to U$ is the canonical map given by

$s\left(B\right)=1\oplus B\oplus \left(B\otimes B\right)\oplus \left(B\otimes B\otimes B\right)\oplus ...\oplus B^{\otimes N}$,

and $p:U\to k$ is some $k$-linear map. (In fact, this follows from the properties of the tensor algebra, because here we are NOT using the algebra structure on our $U$, but we are only using the vector space structure on $U$, and as I said, as a vector space $U$ is just the tensor algebra of $\mathrm{M}_n\left(k\right)$ "cropped" at $N$, which is enough for linearlizing polynomial maps of degree $\leq N$.

Now consider the element $s\left(A\right)\in U$. This element $s\left(A\right)$ is invertible (since $A$ is invertible, so that $A^{\otimes i}$ is invertible for every $i$, and since the multiplication on $U=\bigoplus\limits_{i=0}^N \left(\mathrm{M}_n\left(k\right)\right)^{\otimes i}$ is componentwise), and the algebra $U$ is finite-dimensional (although its dimension is usually quite large). Thus, $s\left(A\right)^{-1}$ lies in the $k$-linear span of the set $\left\lbrace 1,s\left(A\right),\left(s\left(A\right)\right)^2,\left(s\left(A\right)\right)^3,...\right\rbrace$ (because if $u$ is an invertible element of some finite-dimensional $k$-algebra, then $u^{-1}$ lies in the $k$-linear span of the set $\left\lbrace 1,u,u^2,u^3,...\right\rbrace$; this is easily proven using the fact that any element of a finite-dimensional $k$-algebra is algebraic over $k$). Since $s$ is a $k$-algebra homomorphism, we have $\left(s\left(A\right)\right)^i=s\left(A^i\right)$ for all $i$, so that this becomes: The element $s\left(A^{-1}\right)$ lies in the $k$-linear span of the set $\left\lbrace s\left(1\right),s\left(A\right),s\left(A^2\right),s\left(A^3\right),...\right\rbrace$. Since $p$ is a linear map, we can apply $p$ here and obtain: The element $p\left(s\left(A^{-1}\right)\right)$ lies in the $k$-linear span of the set $\left\lbrace p\left(s\left(1\right)\right),p\left(s\left(A\right)\right),p\left(s\left(A^2\right)\right),p\left(s\left(A^3\right)\right),...\right\rbrace$. Now $p\circ s=P$, so this becomes: The element $P\left(A^{-1}\right)$ lies in the $k$-linear span of the set $\left\lbrace P\left(1\right),P\left(A\right),P\left(A^2\right),P\left(A^3\right),...\right\rbrace$. So when $P\left(A^i\right)=0$ for all $i\in\mathbb N$, then $P\left(A^{-1}\right)=0$, qed.