I want to show that the co-efficient of $x^n$ in the function $$ f(x) = (x^2-1)^n $$ when taken the derivative n times is $$ \frac{(2n)!}{n!} $$
To simplify the expression in order to apply Leibnitz formula, I factorize the function $$ f(x) = (x^2-1)^n = (x-1)^n(x+1)^n $$ Now the co-efficient of each term in the Leibnitz formula is a co-efficient of $x^n$, therefore $$ \frac{d^n}{dx^n}(x-1)^n(x+1)^n = \sum_{k=0}^n\frac{n!n!n!}{(n-k)!k!(n-k)!k!} $$
I am unable to prove this equal to $ \frac{(2n)!}{n!} $.
For odd k, the number of terms are even and the formula can be slightly simplified to $$ \frac{d^n}{dx^n}(x-1)^n(x+1)^n = 2\sum_{k=0}^\frac{n-1}{2}\frac{n!n!n!}{(n-k)!k!(n-k)!k!} $$ However this doesn't help me prove the identity.
You need the co-efficient of $x^{2n}$ in $(x^2-1)^n$, because differentiating $x^{2n}$ $n$-times gives you $x^n$. Using the binomial theorem, the coefficient is $1$. On differentiating $n$-times $x^{2n}$, you get $$2n(2n-1)...(2n-(n-1))=\frac{(2n)!}{n!}$$