Prove by induction: $\dfrac{d^{2n}}{dx^{2n}}(x^2-1)^n = (2n)!$ Let $P_n$ be the statement that $\dfrac{d^{2n}}{dx^{2n}}(x^2-1)^n = (2n)!$ 
Base case: n = 0, $\dfrac{d^0}{dx^0}(x^2-1)^0 = 1 = 0!$
Assume $P_m = \dfrac{d^m}{dx^m}(x^2-1)^m = m!$  is true. 
Prove $P_{m+1} = \dfrac{d^{2(m+1)}}{dx^{2(m+1)}}(x^2-1)^{m+1} = [2(m+1)]!$ 
$\dfrac{d^{2(m+1)}}{dx^{2(m+1)}}(x^2-1)^{m+1}$
= $\dfrac{d^{2m}}{dx^{2m}}\left(\dfrac{d^2}{dx^2}(x^2-1)^{m+1}\right)$ 
= $\dfrac{d^{2m}}{dx^{2m}}\left(2x(m)(m+1)(x^2-1)^{m-1}\right)$
= $[\dfrac{d^{2m}}{dx^{2m}}(x^2-1)^m][2x(m)(m+1)(x^2-1)^{-1}]$
From the inductive hypothesis, 
= $(2m)! [2x(m)(m+1)(x^2-1)^{-1}]$ 
I got stuck here, and not sure if I have done correctly thus far?  I did not know how to get to $[2(m+1)]!$. Please advise. Thank you. 
 A: You've made several mistakes. 
Hint: What does the product rule say $\frac{d}{dx} f(x) g(x)$ is equal to?
Now set $ f(x) = x^2 -1, g(x) = (x^2 -1)^m$.   
A: Try the binomial identity
$$
\frac{d^{2n}}{dx^{2n}}(x^{2}+1)(x^{2}+1)^{n-1}=\sum_{i=0}^{2n}{\binom{2n}{i}\frac{d^{i}}{dx^{i}}(x^{2}+1)\frac{d^{2n-i}}{dx^{2n-i}}(x^{2}+1)^{n-1}}
$$
A: Assume:
$$P_m = \dfrac{d^{2m}}{dx^{2m}}(x^2-1)^m = (2m)!$$
Then:
$$P_{m+1}=\dfrac{d^{2m}}{dx^{2m}}\left(\dfrac{d^2}{dx^2}(x^2-1)^{m+1}\right)=
\dfrac{d^{2m}}{dx^{2m}}\left(\dfrac{d}{dx}\left[2x(m+1)(x^2-1)^m\right]\right)=\\
2(m+1)\dfrac{d^{2m}}{dx^{2m}}\left((x^2-1)^m+2\overbrace{x^2}^{x^2-1+1}m(x^2-1)^{m-1}\right)=\\
\color{blue}{2(m+1)\dfrac{d^{2m}}{dx^{2m}}\left((x^2-1)^m+2m(x^2-1+1)(x^2-1)^{m-1}\right)}=\\
\color{blue}{2(m+1)\dfrac{d^{2m}}{dx^{2m}}\left((x^2-1)^m+2m(x^2-1)^{m}+2m(x^2-1)^{m-1}\right)}=\\
2(m+1)\dfrac{d^{2m}}{dx^{2m}}\left((2m+1)(x^2-1)^m+2m(x^2-1)^{m-1}\right)=\\
2(m+1)(2m+1)\dfrac{d^{2m}}{dx^{2m}}(x^2-1)^m+\overbrace{4m(m+1)\dfrac{d^{2m}}{dx^{2m}}(x^2-1)^{m-1}}^{0}=\\
\color{blue}{2(m+1)(2m+1)\dfrac{d^{2m}}{dx^{2m}}(x^2-1)^m+\overbrace{4m(m+1)\dfrac{d^{2m}}{dx^{2m}}P_{2m-2}}^0}=\\
(2m+2)(2m+1)(2m)!=(2(m+1))!$$
$\color{blue}{\text{where $P_{2m-2}$ is a polynomial of degree $2m-2$, whose $2m$-th order derivative is zero.}}$
