Induction principle: $(1+x)^{n+1}>1+(n+1)x$ Given $(1+x)^{n+1}>1+(n+1)x$ for $x>0$ and for $n$ positive integer.
Putting $n=1$ I get $x^2>0$ so the inductive hypothesis is true.
Putting $n=n+1$ I get $(1+x)^{n+2}>1+(n+2)x$. I can write $(1+x)^{n+2}$ as $(1+x)^n(1+x)^2$ and then I don't know how to go on. 
 A: Hint: Multiplying your inequality by $1+x>0$ we get
$$(1+x)^{n+2}>(1+(n+1)x)(1+x)>1+(n+2)x$$ Show this.
The right hand side is:
$$1+nx+x+x+nx^2+x^2>1+nx+2x$$ and this is $$x^2(n+1)>0$$
A: $\textbf{Proof}$
Since $(1+x)^{2}=1+2x+x^2>1+2x$, the inequality holds for $n=1$. Suppose that it holds for $n=k$, then $$\begin{align*}(1+x)^{k+2}&=(1+x)^{k+1}(1+x)\\&>[1+(k+1)x] (1+x)\\&=1+(k+2)x+(k+1)x^2\\&>1+(k+2)x .\end{align*}$$ This shows that the inequality holds for $n=k+1.$ By mathematical induction, it necessarily holds for all $n \in \mathbb{N_+}.$
A: Hint
The best way is write
$$(1+x)^{n+2}=x^{n+1}(x+1)>[1+(n+1)x](x+1)$$
Remmember, you have to get $(1+x)^{n+2}>1+(n+2)x$
Can you finish?
A: We assume that $(1+x)^{n+1}>1+(n+1)x$
We need to prove that $(1+x)^{n+2}>1+(n+2)x$ is true.
From the first line, we have
$(1+x)(1+x)^{n+1}>(1+x)(1+(n+1)x)=1+(n+1)x+x+x^2>1+(n+2)x$
A: No, write
$$(x+1)^{n+2}=(x+1)^{n+1}(x+1)$$ instead of  $$(x+1)^{n+2}=(x+1)^n(x+1)^2.$$ 
Then you can use the induction hypothesis and
$$(x+1)^{n+1}(x+1)>(1+(n+1)x)(1+x)=1+(n+2)x+x^2>1+(n+2)x.$$

In fact it is easier to start with the equivalent statement
$$(x+1)^n>1+nx$$ for $n>1$.
Then
$$(x+1)^{n+1}>(1+nx)(x+1)>1+(n+1)x$$
