Is this equality about derivative of a polynomial valid? Why is $\left(x^2-1\right)\frac{d}{dx}\left(x^2-1\right)^n = 2nx\left(x^2-1\right)^{n-1}$? This is in a textbook and says that its proof is left as an exercise. It seems to be a difficult identity.
I believe this should just be $\left(x^2-1\right)\frac{d}{dx}\left(x^2-1\right)^n = 2nx\left(x^2-1\right)^{n}$ by simple differentiation. 
Which is true?

Just to summarize what various commenters said, the chain rule tells us that
$$(f^n)'=nf^{n-1}f'.$$
If we let $f(x)=x^2-1$, we can see that
$$\frac{d}{dx}\left[\left(x^2-1\right)^n\right]=n\left(x^2-1\right)^{n-1}(2x).$$
Multiplying both sides of the above equation by $x^2-1$, we have, as you correctly assumed:
$$\begin{align}
\left(x^2-1\right)\frac{d}{dx}\left[\left(x^2-1\right)^n\right]&=\left(x^2-1\right)n\left(x^2-1\right)^{n-1}(2x)\\
&=2nx\left(x^2-1\right)^n
\end{align}$$
