Prove the formula: $$\lim_{z\to 0} \frac{d^{2n}}{dz^{2n}}(z^2-1)^{2n}=(-1)^n\left(\frac{(2n)!}{n!}\right)^2.$$
The only way I can conceive to move forward with my limited background is to try induction. Base case obviously checks out, next try to use the assumption the formula works for $n$ to get it for $n+1$: $$\frac{d^{2n}}{dz^{2n}}(\frac{d^2}{dz^2}(z^2-1)^{2n+2})\\ =\frac{d^{2n}}{dz^{2n}}(2z(2n+2)(2n+1)(z^2-1)^{2n}+2(2n+2)(z^2-1)^{2n+1}).$$ Now I am stuck, since if I want to use the formula for $n$ in the first term I can't because of the extra $z$, and in the second term I can't because of the $2n+1$ power term.
Can anyone help please?
Editing again:
In the second term I can actually split of the extra $(z^2-1)$ factor, leading to: $$\frac{d^{2n}}{dz^{2n}}(-1)2(2n+2)(z^2-1)^{2n}.$$
Assuming all the other factors will go to $0$ because of the limit (?), we can compare $$(-1)(4n+4)A_{n}$$ to the original formula for $n$ where I call the right side ($(-1)^n\left(\frac{(2n)!}{n!}\right)^2$) $A_n$. Now substituting $n+1$ for $n$ in $A_n$, leads me to $$(\frac{2n!}{n!})^2(\frac{(2n+2)(2n+1)}{(n+1)})^2,$$ where the second factor does not equal to $(4n+4)$, meaning my approach is the incorrect approach. So I am stuck a bit deeper. Btw, this is a textbook exercise for self study.
Final edit:
Using the binomial theorem as recommended by Gary in the comments made it pretty easy. So the remaining question is solved. What remains unsolved is why I got the inequality in the end when trying induction, so I am doing something wrong. You are welcome to point out what, but of course this warrants a new question. In any case thank you all.
