If $0(1-h)^{n}$ 
Prove that  for $0 < h < 1$ and $n\in\mathbb{N}$,
  $$\frac{1}{1+nh}>(1-h)^{n}.$$

I am trying with induction , but necessary i would be happy for another way to prove this, here what i got :
first we prove that the inequality  correct for $n = 1$ when $0 < h < 1$ .
Second we assume that the inequality correct for $n$ .
then we should prove that the inequality is also correct for $n+1$ .
$$\frac{1}{1+nh+h}>\frac{1-h}{nh+1}\implies
\frac{1}{nh+h+1}>\frac{1-h}{nh+1}$$
How to continue from here ? Or there another way to prove this ?
 A: As an alternative 
$$\frac{1}{1+nh}>(1-h)^{n} \iff \frac1{1-h} >(1+nh)^\frac1n$$
and by Bernoulli's inequality
$$(1+nh)^\frac1n <1+h <\frac1{1-h}$$
indeed
$$1+h<\frac1{1-h}\iff 1-h^2<1$$
which is true.
A: Induction is fine. For the inductive step, note that
$$(1-h)^{n+1}=(1-h)^{n}(1-h)<\frac{1}{1+nh}(1-h)$$
where $1-h>0$. So it remains to show that
$$\frac{1}{1+nh}(1-h)\leq \frac{1}{1+(n+1)h}$$
that is
$$(1+(n+1)h)(1-h)\leq 1+nh$$
or
$$1+nh-\underbrace{(n+1)h^2}_{>0}\leq 1+nh$$
which holds.
A: Note that $0\lt h\lt1$ means we can write $h={x\over1+x}$ with $x\gt0$. Doing so turns the inequality to be proved into
$${1\over1+{nx\over1+x}}\gt\left(1\over1+x\right)^n$$
and this is equivalent to
$$(1+x)^{n+1}\gt1+(n+1)x$$
which is easy enough to prove by induction:
$$(1+x)^{1+1}=1+2x+x^2\gt1+2x=1+(1+1)x$$
and $(1+x)^{n+1}\gt1+(n+1)x$implies
$$\begin{align}
(1+x)^{(n+1)+1}
&=(1+x)(1+x)^{n+1}\\
&\gt(1+x)(1+(n+1)x)\\
&=1+(1+n+1)x+(n+1)x^2\\
&\gt1+((n+1)+1)x
\end{align}$$
Remark: One thing to be aware of is that the strict inequality $1/(1+nh)\gt(-1h)^n$ does not hold when $n=0$, so you really need to be a bit careful in saying "$n\in\mathbb{N}$," since some conventions have $0\in\mathbb{N}$.
A: Another way to prove it is by applying the AM-GM inequality: 
$(1-h)^n(1+nh) < \left(\dfrac{(1-h)+(1-h)+\cdots + (1-h) + (1+nh)}{n+1}\right)^{n+1}= \left(\dfrac{n+1}{n+1}\right)^{n+1}=1\implies \dfrac{1}{1+nh}>(1-h)^n$ .
