Trouble with proving $A$ is an integrally closed domain $\Rightarrow$ $A[t]$ is integrally closed domain

Assume that $f\in k[t]$ is integral over $A[t]$ and $f^n + a_{n-1}(t)f^{n-1} + \cdots + a_1(t)f + a_0(t) = 0$, where $a_i(t)\in A[t]$. Let $m$ be an integer greater than the degree of $f$ and all the degrees of $a_i$. Set $f_1(t)=f(t)-t^m$. If $q(x)=x^n+a_{n-1}(t)x^{n-1}+\cdots+a_1(t)x+a_0(t)$, then $f_1$ is a root of $q_1(x)=q(x+t^m)$. Note that $q_1(x)=x^n+b_{n-1}(t)x^{n-1}+\cdots+b_1(t)x+b_0(t)$ and $b_0(t)=q(t^m)$ is monic. Since $f_1(f_1^{n-1}+b_{n-1}f_1^{n-2}+\cdots +b_1(t))=-b_0(t)$ and $f_1$ and $b_0$ are monic, it follows that $f_1^{n-1}+b_{n-1}f_1^{n-2}+\cdots +b_1(t)$ is also monic, and now you can apply Gauss Lemma.

Here is another way of doing it. We denote the field of fractions of $A$ by $K$. Now we first note that $K[t]$ is integrally closed in $K(t)$ (this follows readily from the fact that UFD's are integrally closed). So now all we have to show is that $A[t]$ is integrally closed in $K[t]$. To do this we choose a polynomial $f \in K[t]$ that is integral over $A[t]$ such that $f$ has degree $n$ in the variable $t$. We also assume that we have shown the result (that is, if $f \in K[t]$ is integral over $A[t]$ then it is in $A[t]$) for all $f$ with degree less than $n$. We first recall the mantra "integral + finite type = finite". Now we consider the extension

$ A[t] \subset (A[t])[f] $

It is finite as it is integral and finite type. Now as one can check, the ring of leading coefficients of polynomials in $(A[t])[f]$, denoted by $L$, is finite over $A$. In particular, if

$ f = k_nt^n+\dots+k_0 $

we have that $A[k_n] \subset L$, and hence $k_n$ is integral over $A$. Since $A$ is integrally closed in its field of fractions, we have that $k_n \in A$. We now note that $f-k_nt^m$ is integral over $A[t]$ and has degree less than $n$. We finish the proof by noting that the result for $n=0$ is equivalent to the integral closure of $A$ in $K$.