Let $x>-1$ , then statement $P\left(n\right):(1+x{)}^n>1+nx$ is true for 

For $n=2,$

$P\left(2\right):(1+x{)}^2=1+2x+x^2>1+2x$ as $x\ne 0$

Assume that $P\left(k\right):(1+x{)}^k>1+kx$

For some   $k\in \mathbf{N},k>1$

As $x>-1$ multiplying both the sides of $\left(1\right)$ by we $1 + x,$ get

$\begin{array}{l}(1+x{)}^{k+1}>(1+kx)(1+x)\\ =1+(k+1)x+k{x}^2\\ >1+(k+1)x \left[\because k{x}^2>0\right]\end{array}$

Thus $, P(k + 1)$ is true.

By the principle of mathematical induction $P\left(n\right)$  is true for 

all $n>1$provided $x\ne 0$