Suppose that we have the expression $$ xr_n(x)=b_nr_{n+1}(x)+a_nr_n(x)+b_{n-1}r_{n-1}(x),\quad n\geq 0 $$ where $a_n\in \mathbb{R}$, $b_n>0$ for $n\geq 0$ and $b_{-1}=1$, with the initial conditions $r_0(x)=1$ and $r_{-1}(x)=0$.
From here, I can see that $r_n$ is a (real) polynomial of degree $n$ (by induction). Then, we can write $r_n(x)=\sum_{k=0}^nc_{k,n}x^k$ for some coefficients $c_{k,n}$. How can one deduce from the expression above that $$ c_{n,n}=1/(b_0b_1\dots b_{n-1})? $$
Use induction. Notice that $\deg a_n r_n(x) \leq n$ and $\deg b_{n-1}r_{n-1}(x) \leq n-1$ (building on your result that $r_n(x)$ has degree $n$). So when looking at coefficient of $x^{n+1}$ in the recurrence expression, the only terms left are $xr_n(x)$ and $b_nr_{n+1}(x)$. The coefficient of $x^{n+1}$ in $xr_n(x)$ is the coefficient of $x^n$ in $r_n(x)$, which by induction hypothesis is $1/(b_0b_1\dots b_{n-1})$. This must be equal to coefficient of $x^{n+1}$ in $b_nr_{n+1}(x)$, and hence coefficient of $x^{n+1}$ in $r_{n+1}(x)$ is $(1/(b_0b_1\dots b_{n-1}))/b_n=1/(b_0b_1\dots b_{n})$, which was to be proven.
