Chapter 12.4, Problem 2P

Show that $P_l(1)=1,$ with $P_1(x)$ given by (4.1), in the following way. Factor ${\left(x^2-1\right)}^l$ into ${(x+1)}^l{(x-1)}^1$ and differentiate the product $l$ times by Leibniz' rule. Without writing out very many terms you should see that every term but one contains the factor $x-1$ and so becomes zero when $x=1.$ Use this to evaluate $P_l(x)$ in (4.1) when $x=1$ to get $P(1)=1$.