Prove that $$\int_1^{\frac{1+z}{1-z}} \frac{d^n}{dz^n} (z^2-1)^n= -\frac{2(z-1)^n}{(n+1)! }\frac{d^n}{dz^n} (\frac{z}{z-1})^{n+1} $$
Here are some attempts.
$\frac{d^n}{dz^n} (z^2-1)^n=2^n n! L_n(z),$ where $L_n(z)$ is the Legendre polynomials. I tried to use following formula for Legendre polynomials $$\frac{x^2-1}{n} \frac{d}{dx}L_n(x)=xL_n(x)-L_{n-1}(x).$$ But I get tripped up on it.
Since $\frac{d^n}{dz^n} (z^2-1)^n$ is Rodrigues' formula of Legendre polynomials, we use the following formula from https://en.wikipedia.org/wiki/Legendre_polynomials.
$L(z)=\sum_{i=0}^n {{n \choose i} {-n-1 \choose i}(\frac{1-x}{2})^{i}}.$
Using the property ${n \choose i}=(-1)^i{-n+i-1 \choose i}$ and leting $k:=n-i$ we have
$L_n(x)=\sum_{k=0}^n {{n \choose k} {2n-k \choose n}(\frac{x-1}{2})^{n-k}}. $
Thus we have
$\int_1^{\frac{1+z}{1-z}} \frac{d^n}{dz^n} (z^2-1)^n= \int_1^{\frac{1+z}{1-z}}L_n(z) dx=\int^{\frac {z+1}{1-z}}_{1} \sum_{k=0}^n {{n \choose k} {2n-k \choose n}(\frac{x-1}{2})^{n-k}} dx= \\= \left.\sum_{k=0}^n {n \choose k} {2n-k \choose n}\frac{(x-1)^{n-k+1}}{2^{n-k}(n-k+1)}\right|_1 ^{\frac {z+1}{1-z}}= \\=\frac{1}{n+1} \sum_{k=0}^n {n+1 \choose k} {2n-k \choose n}(1-\frac{z+1}{1-z})^{n-k}\frac{1}{2^{n-k+1}}-\frac{1}{n+1} \sum_{k=0}^n {n+1 \choose k} {2n-k \choose n}(0)^{n-k+1}\frac{1}{2^{n-k}}= \\=\frac{1}{n+1} \sum_{k=0}^n {n+1 \choose k} {2n-k \choose n}\frac{(2z)^{n-k+1}}{2^{n-k}(1-z)^{n-k+1}}= \\=2\frac{(z-1)^{n} (-1)^n}{n+1} \sum_{k=0}^n {n+1 \choose k} {2n-k \choose n}z^{n+1-k}(1-z)^{-2n+k-1}.$
Appling the General Leibniz Rule for product differentiation for $\frac{d^n}{dz^n}\left(\frac{z}{z-1}\right)^{n+1}$ we have $-\frac{2(z-1)^{n}}{(n+1)!} \frac{d^n}{dz^n}\left(\frac{z}{z-1}\right)^{n+1}= =-\frac{2(z-1)^{n}}{(n+1)!} \sum_{k=0}^n {n \choose k} \frac{d^k }{dz^k}z^{n+1}\frac{d^{n-k} }{dz^{n-k}}\frac{1}{(z-1)^{n+1}}=\\ -\frac{2(z-1)^{n}}{n+1} \sum_{k=0}^n {n+1 \choose k} {2n-k \choose n}\frac{(n+1-k)!}{(n+1)!}\frac{d^k z^{n+1}}{dz^k}\frac{n!}{(2n+k)!}\frac{d^{n-k} (z-1)^{-(n+1)}}{dz^{n-k}}= \\ 2\frac{(z-1)^{n} (-1)^n}{n+1} \sum_{k=0}^n {n+1 \choose k} {2n-k \choose n}z^{n+1-k}(1-z)^{-2n+k-1}.$
There is another proof. But it uses a little-known identity.