Derivative of degree k for $f(t)$ $=$ $1 \over {1 + t}$ 
Given
$f: \Bbb R \setminus \{-1\} \rightarrow \Bbb R$,
$f(t)$ $=$ $1 \over {1 + t}$,
I would like to calculate the derivative of degree $k$.

Approach
First, we try to examine if there is a certain pattern that can be observed.
Applying the quotient rule a few times, we receive:
$f'(t)$ $=$ $-1 \over {(1 + t)^2}$,
$f''(t)$ $=$ $2 \over {(1 + t)^3}$,
$f'''(t)$ $=$ $-6 \over {(1 + t)^4}$,
...
The structure of the derivatives becomes quite obvious. Therefore, we claim:
$f^k$ $=$ $(-1)^k$ $k! \over (1 + t)^{k+1}$
Proof by induction
For $k = 0$ we get
$(-1)^0$ $0! \over (1 + t)^{0 + 1}$ $=$ $1 \over (1 + t)$
Now, assume the statement has already been proven for $k = n.$ We have to show that it also holds for $k = n+1.$
$f^{(n+1)}$ $=$ $(-1)^{n+1}$ $(n+1)! \over (1 + t)^{n+2}$ $=$ $(-1)^n$ $n! \over (1 + t)^{n+1}$ $(-1)$ $n+1 \over (1 + t)$ $= f^n$ $(-1)$ $n+1 \over (1 + t)$ $=$ ...,
and this is where I'm stuck. I just have to show that the expression on the right side is the first derivative of the function, right? But this doesn't work out, I guess.
 A: If $$f^{(n)}(t)=(-1)^n\frac{n!}{(1+t)^{n+1}}$$
then
$$f^{(n+1)}(t)=\left(f^{(n)}\right)'(t)=(-1)^nn![-(n+1)](1+t)^{-(n+2)}=\left(-1\right)^{n+1}\frac{(n+1)!}{(1+t)^{n+2}}$$
A: Assume the statement has been already proven for $k = n.$ 
That is $$f^{n}=(-1)^n\frac{n!}{(1+t)^{n+1}}$$
Now differentiate $f^n$ , then you will get 
$$\frac{df^{n}}{dt}=(-1)^n \cdot n!\cdot \frac{d}{dt}\frac{1}{(1+t)^{n+1}}$$
$$\frac{df^{n}}{dt}=(-1)^n \cdot n!\cdot \frac{(-1)(n+1)}{(1+t)^{n+2}}$$
Thus $$f^{n+1}=(-1)^{n+1} \cdot (n+1)!\cdot \frac{1}{(1+t)^{n+2}}$$
A: In the induction step, you assume that the formula is true for $k = n$. But when you are showing for $k = n+1$ you are saying that $f^{(n+1)}(t) = (-1)^{n+1} \frac{(n+1)!}{(1+t)^{n+2}}$, which is exactly what you want to prove! Your reasoning is circular.
What you know is that $f^{(n)}(t) = (-1)^n \frac{n!}{(1+t)^{n+1}}$. Then $f^{(n+1)}(t) = (f^{(n)}(t))' = \left( (-1)^n \frac{n!}{(1+t)^{n+1}} \right)'$. I leave the rest to you.
A: When you do the inductive step you have to start from the hypothesis that the statement holds for $k=n$ and then conclude that it holds for $k=n+1$. In your proof you already assume the conclusion. Instead suppose that the statement holds for $k=n$ and note that 
$$
\begin{align}
f^{(n+1)}(t)
&=\frac{d}{dt}\left( (-1)^n \frac{n!}{(1+t)^{n+1}}\right)\\
&=(-1)^n \times(1-n)\times n!(1+t)^{-(n+2)}\\
&=(-1)^{n+1} \frac{(n+1)!}{(1+t)^{n+2}}
\end{align}
$$
where we have used the power rule in the second line and the inductive hypothesis in the first line.
