I am struggling to see how P_{k}(x)= [1/(2^{k}(k!))][(d^{k}((x^{2}-1)^{k})/dx^{k}] = [1/(2^{k})][2k choose k][x^{k}] + (terms of lower order)
Help please?
$\displaystyle (x^2-1)^k$ is a polynomial of degree $\displaystyle 2k$; when you take the $\displaystyle k$-th derivative it's a polynomial of degree $\displaystyle k$. Now, look at the leading term.