How to change the variables in $\frac{1}{(z+1)^{n+1}}=\frac{(-1)^n}{n!}\frac{d^n}{dz^n}\left( \frac{1}{z+1}\right)$? We know that 
$$\frac{1}{(z+1)^{n+1}}=\frac{(-1)^n}{n!}\frac{d^n}{dz^n}\left( \frac{1}{z+1}\right).$$
If we put $z=e^t$ how can we change the variables above equation?
I know that $z\frac{d}{dz}=\frac{d}{dt}$. So how can we write $\frac{d^n}{dz^n}\left( \frac{1}{z+1}\right)$?
Is $z^n \frac{d^n}{dz^n}=\frac{d^n}{dt^n}$?

In general, if $f(z)$ is $n$ times differentiable, then we have
$$\left(z\frac{d}{dz}\right)^n\{f(z)\}=\sum_{k=1}^n  \begin{Bmatrix} n \\ k \end{Bmatrix}z^k\frac{d^k\,f(z)}{dz^k}$$
where $\begin{Bmatrix} n \\ k \end{Bmatrix}=\frac1{k!}\sum_{j=0}^k (-1)^{k-j}\binom{k}{j}\,j^n$ are the Stirling Numbers of the Second Kind.
For $f(z)=\frac1{z+1}$, we find
$$\begin{align}
\left(z\frac{d}{dz}\right)^n\left(\frac{1}{1+z}\right)=\sum_{k=1}^n  \begin{Bmatrix} n \\ k \end{Bmatrix}z^k\frac{(-1)^k\,k!}{(1+z)^{k+1}}
\end{align}$$
Letting $z=e^t$, we find that 
$$\begin{align}
\frac{d^n}{dt^n}\left(\frac{1}{1+e^t}\right)&=\frac{1}{1+e^t}\sum_{k=1}^n  \begin{Bmatrix} n \\ k \end{Bmatrix}(-1)^k\,k! \left(\frac{e^t}{1+e^t}\right)^k\\\\
&=\frac1{1+e^t}\sum_{k=1}^n\sum_{j=0}^k (-1)^j\binom{k}{j}j^n\left(\frac{1}{1+e^{-t}}\right)^k
\end{align}$$
