This is the final part of a calculation that proceeds from this previous question. Here is almost a word for word copy of the textbook question:
Use the recursion relations below (for the $N_n(x)$'s and the $J_n(x)$'s): $$\frac{\mathrm{d}}{\mathrm{d}x}\left[x^nJ_n(x)\right]=x^nJ_{n-1}(x)\tag{1}$$ $$\frac{\mathrm{d}}{\mathrm{d}x}\left[x^{-n}J_n(x)\right]=-x^{-n}J_{n+1}(x)\tag{2}$$ $$J_{n-1}(x)+J_{n+1}(x)=\frac{2n}{x}J_n(x)\tag{3}$$ $$J_{n-1}(x)-J_{n+1}(x)=2J_{n}^{\prime}(x)\tag{4}$$ $$J_{n}^{\prime}(x)=-\frac{n}{x}J_{n}(x)+J_{n-1}(x)=\frac{n}{x}J_n(x)-J_{n+1}(x)\tag{5}$$ where $$N_n(x)=\frac{\cos(n\pi )J_n(x)-J_{-n}(x)}{\sin(n\pi )}\tag{6}$$ and $$\begin{align}J_n(x)N_n^{\prime}(x)-J_{n}^{\prime}(x)N_n(x)=\frac{J_{n}^{\prime}(x)J_{-n}(x)-J_n(x)J_{-n}^{\prime}(x)}{\sin(n\pi)}=\frac{2}{\pi x}\end{align}\qquad\uparrow\text{A Result that follows from this previous question}\tag{7}$$
to show that $$\fbox{$J_n(x)N_{n+1}(x)-J_{n+1}(x)N_n(x)=-\frac{2}{\pi x}$}\tag{8}$$$\bbox[#AFA]{\text{HINT: Do it first for $n = 0$; then use the result in proving the $n = 1$ case, and so on.}}$
Firstly, rewriting $(5)$ in terms of $N_n(x)$'s:
$$N_{n}^{\prime}(x)=\frac{n}{x}N_{n}(x)-N_{n+1}(x)$$ Starting from the leftmost side of equation $(7)$ and substituting expressions for $N_n^{\prime}(x)$ and $J_n^{\prime}(x)$:
$$\begin{align}\require{enclose}J_n(x)N_n^{\prime}(x)-J_{n}^{\prime}(x)N_n(x)&=J_n(x)\left[\frac{n}{x}N_{n}(x)-N_{n+1}(x)\right]-\left[\frac{n}{x}J_n(x)-J_{n+1}(x)\right]N_n(x)\\&=\enclose{downdiagonalstrike}{\frac{n}{x}J_n(x)N_{n}(x)}-J_n(x)N_{n+1}(x)\enclose{downdiagonalstrike}{-\frac{n}{x}J_n(x)N_{n}(x)}+J_{n+1}(x)N_n(x)\\&=-\left[J_n(x)N_{n+1}(x)-J_{n+1}(x)N_n(x)\right]\\&\end{align}$$ and so by equation $(7)$ $$J_n(x)N_{n+1}(x)-J_{n+1}(x)N_n(x)=-\left[J_n(x)N_n^{\prime}(x)-J_{n}^{\prime}(x)N_n(x)\right]=-\frac{2}{\pi x}$$ as required. $\large\fbox{}$
So what is my question then? Well it's simple really; unless I've made a mistake or got very lucky I managed to prove equation $(8)$ without the use of the HINT highlighted in green (which was a $\color{red}{\mathrm{requirement}}$ of the question).
This leads me to believe I have done something wrong or my calculation was a pure fluke. So my question is; Could someone please verify whether or not this is a lucky coincidence, and if so, is there some way to employ the green HINT?
Thank you.