$ 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 !

To prove
$x\in \left]0,{1\over n-1} \right[ \implies
 1+nx \le (1+x)^n \le {1+x\over 1-(n-1)x}
$.
The first inequality
is just Bernoulli's inequality,
is true for all $x \ge 0$,
and is easily proved by induction:
It is true for $n=1$.
If it is true for $n$,
then
$(1+x)^{n+1}
=(1+x)(1+x)^n
\ge (1+x)(1+nx)
=1+(n+1)x+nx^2
\ge 1+(n+1)x
$.
The second inequality 
is the same as
$(1-(n-1)x)(1+x)^{n-1} \le 1
$.
If $n = 1$,
this is
$1 \le 1$,
which is true.
Suppose it is true for $n$.
Then,
if $0 < x < \frac1{n}$,
$\begin{array}\\
(1-nx)(1+x)^{n}
&=(1-(n-1+1)x)(1+x)(1+x)^{n-1}\\
&=(1+x)(1-(n-1+1)x)(1+x)^{n-1}\\
&=(1+x)((1-(n-1)x)(1+x)^{n-1}-x(1+x)^{n-1})\\
&=(1+x)((1-(n-1)x)(1+x)^{n-1}-x(1+x)^{n-1})\\
&\le(1+x)(1-x(1+x)^{n-1})
\qquad\text{by the induction hypothesis}\\
&\le(1+x)(1-x)
\qquad\text{since }(1-x)^{n-1} \ge 1\\
&=1-x^2\\
&\le1\\
\end{array}
$
