$ x\in \left[0,{1\over n-1} \right] \to 1+nx \le (1+x)^n \le {1+x\over 1-(n-1)x}$ (Homework assignement) This is about a homework I have to do. I don't want the straight answer, just the hint that may help me start on this. To give you context, we're now studying integrals.
Now here is the question :

Prove : $ x\in \left[0,{1\over n-1} \right] \to 1+nx \le (1+x)^n \le {1+x\over 1-(n-1)x}$

The exercice suggest using what I can only poorly translate to "Inequality of finite increasing" and that states :

Let $f$ be a function continuous on $[a,b], a<b$ and differentiable on $[a,b]$.
$\exists M \in \mathbb{R}, \forall x \in [a,b], f'(x)\le M \to f(b)-f(a) \le M(b-a)$

I tried to apply this to $f(x)=(1+x)^n$ but to no avail.
Any input will be greatly apreciated, Thanks !

For the inequality  $1+nx\leq (1+x)^n $, just expand using the binomial theorem and notice that all terms are positive. 
The other inequality, after some manipulations (note that all terms are positive) looks like $$1-(n-1)x\leq (1+x)^{-(n-1)}. $$ 
  Consider the function $$f (x)=(1+x)^{-(n-1)}+(n-1)x-1.$$ We have $f(0)=0$, and $$f'(x)=-(n-1)(1+x)^{-n}+n-1=(n-1)(1-(1+x)^{-n})>0,$$since $(1+x)^{-n}<1$. We have, then, that $f (0)=0$ and $f $ is increasing.
