= Solution
Evaluate $R_n$ at $x=1$ using $(x^2-1)^n=(x-1)^n(x+1)^n$. In the <Leibniz rule> for the $n$th <derivative>, every term vanishes at $1$ except the term putting all $n$ <derivatives> on $(x-1)^n$. Therefore $R_n(1)=n!2^n$. Combining this with $P_n(1)=1$ gives the normalisation in the <Rodrigues' formula>:
$$
\boxed{\alpha_n=\frac1{2^n n!},\qquad P_n(x)=\frac1{2^n n!}\frac{d^n}{dx^n}(x^2-1)^n}.
$$
For $n=0$, the formula gives $\alpha_0=1$, as required. As an independent check, the leading term of $R_n$ is $(2n)!x^n/n!$, so the resulting <leading coefficient> of $P_n$ is $(2n)!/(2^n(n!)^2)$, consistent with the <Legendre polynomial recurrence relation>.
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