I'm trying to construct the Legendre polynomials from the differential equation. As is done in this set of lecture notes, I can get an expression for the coefficient $c_{l-2k}$ in terms of $c_l$: $$c_{l-2k} = (-1)^k \frac{l!l!}{(2l)!}\frac{(2l-2k)!}{k!(l-k)!(l-2k)!} c_l$$ From this, the lecture notes say that we must use the fact that $P_l(1) = 1$ to show that $$c_l = 2^{-l}\binom{2l}{l}$$ and then plug this into the previous formula to simplify the series formula into the standard expression $$P_l(x) = \sum_{k=0}^{\lfloor l/2\rfloor}c_{l-2k}x^{l-2k} = \sum_{k=0}^{\lfloor l/2\rfloor} \left[\frac{(-1)^k}{2^l}\binom{2l-2k}{l}\binom{l}{k}\right]x^{l-2k}$$ My question is: how do we arrive at this expression for $c_l$? All of the explanations I've seen so far (including but not limited to Boas's mathematical physics textbook, Balakrishnan's mathematical physics textbook, and several sets of lecture notes available on the internet) either just assume this formula or define the Legendre polynomials using the Rodrigues formula or the generating function (which I'd rather not do because the construction of the Rodrigues formula relies on having a series formula in the first place, and it's not obvious to me where the generating function is derived from, or why it works). The source the lecture notes cite as 'proof' of the expression for $c_l$ is in Portuguese and (from my loose attempts using Google translate) also just outright state this expression without proof.
By rearranging the normalisation condition a little bit, I can reduce the problem to proving that $$\sum_{k=0}^{\lfloor l/2 \rfloor} (-1)^k \frac{(2l-2k)!}{k!(l-k)!(l-2k)!} = 2^l$$ but I have no idea how I would go about proving this. I've tried using induction, and the base case is easy enough to prove, but I don't know how to show this for the recursive case; the expression I get is very convoluted.
