 Solution

Write $v=(x+1{)}^l(x-1{)}^l$. In the Leibniz rule for its $l$th derivative at one, any term with fewer than $l$ derivatives on $(x-1{)}^l$ vanishes. The sole surviving term leaves $(x+1{)}^l$ undifferentiated and differentiates the second factor $l$ times, givingThe normalization in Rodrigues' formula consequently givesThe same argument covers $l=0$, with both zeroth powers equal to one.