2n-th derivation of Legendre Polynomial

Question

Let $u_n(x)=(x^2-1)^n$ Show that $\frac{(d^{2n}u_n(x)} {dx^{2n}} = 2n!$

$(x^2-1)^n = (x-1)^n(x+1)^n $ and then a should use Leibnitz formula. I thought if I write Leibnitz formula as a binomial I can get it but I couldn’t. Could you please help?

Thanks

Answer 1

Much simpler: note that $u_n(x)=(x^2-1)^n$ is a monic polynomial of degree $2n$ in $x$, so taking $2n$-th derivative gives $(2n)!$.

And no, $u_n$ itself is not the Legendre polynomial. which is $u_n^{(n)}$ up to a constant factor.