Polynomial ring over integrally closed subring is integrally closed in polynomial ring

Statement
Suppose $$R$$ is an fact about::integrally closed subring of a commutative unital ring $$S$$. Then, the polynomial ring $$R[x]$$ is an integrally closed subring of $$S[x]$$.

Proof
Given: A ring $$S$$, an integrally closed subring $$R$$.

To prove: $$R[x]$$ is an integrally closed subring of $$S[x]$$.

Proof: Suppose $$f \in S[x]$$ satisfies a monic polynomial over $$R[x]$$:

$$f^n + g_{n-1}f^{n-1} + g_{n-2}f^{n-2} + \dots + g_0 = 0$$.

Here, $$g_i \in R[x]$$ for all math>i. Write $$f = h + x^r$$, for $$r$$ greater than the maximum of the degrees of the $$g_i$$. Note that $$f \in R[x]$$ if and only if $$h \in R[x]$$, so it suffices to show that $$h \in R[x]$$.

$$h$$ satisfies the monic polynomial:

$$(h + x^r)^n + g_{n-1}(h + x^r)^{n-1} + g_{n-2}(h + x^r)^{n-2} + \dots + g_0 = 0$$.

The constant term of this, viewed as a polynomial in $$h$$, is:

$$(x^r)^n + g_{n-1}(x^r)^{n-1} + \dots + g_0$$.

Since all the $$g_i$$ are in $$R[x]$$, this whole constant term is an element of $$R[x]$$.