First I took the Legendre equation: $$(1-x^2)\frac{d^2P_n(x)}{dx^2}-2x\frac{dP_n(x)}{dx}+n(n+1)P_n(x)=0$$ Then I wrote: $$P_n(x)=\sum_{k=0}^{n}a_{n, k} x^k$$ Where $a_{n, k}$ just gives the coefficient of $x^k$ in $P_n(x)$, now using these two, I did some algebra, getting: $$\sum_{k=2}^nk(k-1)a_{n, k}x^{k-2}=\sum_{k=0}^{n}(k+n+1)(k-n)a_{n, k}x^k$$ From this after comparing coefficients I got that $a_{n, n-1}=0$, and: $$(k+n+1)(k-n)a_{n, k}=(k+1)(k+2)a_{n, k+2}$$ So, $a_{n, n-1-2k}=0$ for $k\in \mathbb{N}$
Now I let: $$f=n-2\left\lfloor{\frac{n}{2}}\right\rfloor$$ So, I get: $$P_n(x)=a_{n,f}x^f+\sum_{k=1}^{\left\lfloor{\frac{n}{2}}\right\rfloor}a_{n, f}\left(\prod_{l=0}^{k-1}\frac{(f+n+2l+1)(f-n+2l)}{(f+2l+1)(f+2l+2)}\right)x^{f+2k}$$ Now, I want to know how can I simplify the product, and also normalise this by finding the value of $a_{n, f}$ given that $P_n(1)=1$