Nice way of showing the following equality? Is there a good way of showing that ${d^n\over dx^n}(x^2-1)^n|_{x=1}=2^nn!$?
I have tried  binomial expanding the thing then differntiate term-by-term, which seems a bit clumsy. Perhaps there's a closed form for ${d^n\over dx^n}(x^2-1)^n$? Thanks.
 A: By applying the product rule repeatedly, we get
$$
\frac{\mathrm{d}^n}{\mathrm{d}x^n}(fg)=\sum_{k=0}^n\binom{n}{k}f^{(n-k)}g^{(k)}
$$
let $f=(x-1)^n$ and $g=(x+1)^n$, then at $x=1$, all terms but the $\binom{n}{0}f^{(n)}g^{(0)}$ term are annihilated by $f^{(n-k)}(1)$. Since $f^{(n)}(1)=n!$ and $g^{(0)}(1)=2^n$, we get that $\frac{\mathrm{d}^n}{\mathrm{d}x^n}(x^2-1)^n=2^nn!$
A: Let $f(t) = ((1+t)^2 - 1)^n$.  You want $f^{(n)}(0)$.  Now from the Maclaurin series $f(t) = \sum_{j=0}^\infty \frac{f^{(j)}(0)}{j!} t^j$, $f^{(n)}(0)/n!$ is the coefficient of $t^n$ in 
$f(t)$.  But $f(t) = (2t+t^2)^n = (2 + t)^n t^n$, so the coefficient of $t^n$ there is $2^n$. 
A: I'm not sure how to make it rigorous, but when you take the $n$ derivatives, you need to apply it to the $(x^2-1)^n$ term every time, not to the powers of $2x$ that come from differentiating the inside.  Otherwise you will have at least one factor of $x^2-1$ left, which will give $0$.  Each time you differentiate you get a factor $2x$ times the exponent.  The exponent decreases from $n$ down to $1$, giving the $n!$ term.  You wind up with $(2x)^n n! + (x^2-1)(stuff)$, which evaluated at $x=1$ gives $2^nn!$
